E
Sue Pemberton
Series Editor: Julian Gilbey
PL
Cambridge International
AS & A Level Mathematics:
SA
M
Pure Mathematics 1
Coursebook
Original material © Cambridge University Press 2017
Contents
Contents
Series introduction
vi
How to use this book
viii
1 Quadratics
E
x
Acknowledgements
1
1.1 Solving quadratic equations by factorisation
1.2 Completing the square
3
6
1.3 The quadratic formula
PL
10
1.4 Solving simultaneous equations (one linear and one quadratic)
11
1.5 Solving more complex quadratic equations
15
1.6 Maximum and minimum values of a quadratic function
17
1.7 Solving quadratic inequalities
21
1.8 The number of roots of a quadratic equation
24
1.9 Intersection of a line and a quadratic curve
27
End-of-chapter review exercise 1
31
SA
M
2 Functions
33
2.1 Definition of a function
34
2.2 Composite functions
39
2.3 Inverse functions
43
2.4 The graph of a function and its inverse
48
2.5 Transformations of functions
51
2.6 Reflections
55
2.7 Stretches
57
2.8 Combined transformations
59
End-of-chapter review exercise 2
67
3 Coordinate geometry
70
3.1 Length of a line segment and mid-point
72
3.2 Parallel and perpendicular lines
75
3.3 Equations of straight lines
78
3.4 The equation of a circle
82
Original material © Cambridge University Press 2017
iii
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
3.5 Problems involving intersections of lines and circles
88
End-of-chapter review exercise 3
92
95
4 Circular measure
99
4.1 Radians
101
4.2 Length of an arc
E
Cross-topic revision exercise 1
104
4.3 Area of a sector
107
End-of-chapter review exercise 4
116
PL
5 Trigonometry
112
5.1 Angles between 0° and 90° 118
5.2 The general definition of an angle
121
5.3 Trigonometric ratios of general angles
123
5.4 Graphs of trigonometric functions
127
5.5 Inverse trigonometric functions
136
5.6 Trigonometric equations
140
5.7 Trigonometric identities
145
5.8 Further trigonometric equations
149
End-of-chapter review exercise 5
153
SA
M
iv
6 Series
155
6.1 Binomial expansion of ( a + b ) n 156
6.2 Binomial coefficients
160
6.3 Arithmetic progressions
166
6.4 Geometric progressions
171
6.5 Infinite geometric series
175
6.6 Further arithmetic and geometric series
180
End-of-chapter review exercise 6
183
Cross-topic revision exercise 2
186
7 Differentiation
190
7.1 Derivatives and gradient functions
191
7.2 The chain rule
198
7.3 Tangents and normals
201
7.4 Second derivatives
205
End-of-chapter review exercise 7
209
Original material © Cambridge University Press 2017
Contents
8 Further differentiation
211
213
8.2 Stationary points
216
8.3 Practical maximum and minimum problems
221
8.4 Rates of change
227
8.5 Practical applications of connected rates of change
230
End-of-chapter review exercise 8
235
9 Integration
E
8.1 Increasing and decreasing functions
238
9.1 Integration as the reverse of differentiation
239
9.2 Finding the constant of integration
244
9.3 Integration of expressions of the form ( ax + b ) 247
PL
n
9.4 Further indefinite integration
249
9.5 Definite integration
250
9.6 Area under a curve
253
9.7 Area bounded by a curve and a line or by two curves
260
9.8 Improper integrals
264
9.9 Volumes of revolution
268
End-of-chapter review exercise 9
276
280
Practice exam-style paper
284
SA
M
Cross-topic revision exercise 3
Answers286
Glossary317
Index319
Original material © Cambridge University Press 2017
v
E
PL
SA
M
Chapter 1
Quadratics
In this chapter you will learn how to:
■
■
■
■
■
■
carry out the process of completing the square for a quadratic polynomial ax 2 + bx + c and use
a completed square form
find the discriminant of a quadratic polynomial ax 2 + bx + c and use the discriminant
solve quadratic equations, and quadratic inequalities, in one unknown
solve by substitution a pair of simultaneous equations of which one is linear and one is quadratic
recognise and solve equations in x that are quadratic in some function of x
understand the relationship between a graph of a quadratic function and its associated algebraic
equation, and use the relationship between points of intersection of graphs and solutions of
equations.
Original material © Cambridge University Press 2017
1
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
PREREQUISITE KNOWLEDGE
Where it comes from
What you should be able to do
Check your skills
IGCSE / O Level Mathematics
Solve quadratic equations by
factorising.
1 Solve.
a x 2 + x − 12 = 0
b x 2 − 6x + 9 = 0
IGCSE / O Level Mathematics
Solve linear inequalities.
E
c 3x 2 − 17 x − 6 = 0
2 Solve.
a 5x − 8 . 2
b 3 − 2x ø 7
Solve simultaneous linear
equations.
3 Solve.
PL
IGCSE / O Level Mathematics
a 2 x + 3 y = 13
7 x − 5 y = −1
b 2 x − 7 y = 31
3x + 5 y = −31
IGCSE / O Level Additional
Mathematics
4 Simplify.
a
20
b ( 5 )2
SA
M
2
Carry out simple manipulation
of surds.
c
8
2
Why do we study quadratics?
At IGCSE / O Level, you will have learnt about straight-line graphs and their properties.
They arise in the world around you. For example, a cell phone contract might involve a
fixed monthly charge and then a certain cost per minute for calls: the monthly cost, y, is
then given as y = mx + c where c is the fixed monthly charge, m is the cost per minute and
x is the number of minutes used.
Quadratic functions are of the form y = ax 2 + bx + c (where a ≠ 0) and they have
interesting properties that make them behave very differently from linear functions.
A quadratic function has a maximum or a minimum value, and its graph has interesting
symmetry. Studying quadratics offers a route into thinking about more complicated
functions such as y = 7 x5 − 4x 4 + x 2 + x + 3.
You will have plotted graphs of quadratics such as y = 10 − x 2 before starting your
A Level course. These are most familiar as the shape of the path of a ball as it travels
through the air (called its trajectory). Discovering that the trajectory is a quadratic was one
of Galileo’s major successes in the early 17th century. He also discovered that the vertical
motion of a ball thrown straight upwards can be modelled by a quadratic, as you will learn
if you go on to study the Mechanics component.
Original material © Cambridge University Press 2017
WEB LINK
Explore the
Quadratics resource
on the Underground
Mathematics site
(www.underground
mathematics.org).
Chapter 1: Quadratics
1.1 Solving quadratic equations by factorisation
You already know the factorisation method and the quadratic formula method to solve
quadratic equations algebraically.
This section consolidates and builds on your previous work on solving quadratic equations
by factorisation.
E
EXPLORE 1.1
2 x 2 + 3x − 5 = ( x − 1)( x − 2)
This is Rosa’s solution to the previous equation:
( x − 1)(2x + 5) = ( x − 1)( x − 2)
PL
Factorise the left hand side:
2x + 5 = x − 2
Divide both sides by ( x − 1):
x = −7
Rearrange:
Discuss her solution with your classmates and explain why her solution is not fully correct.
Now solve the equation correctly.
WORKED EXAMPLE 1.1
SA
M
Solve
a 6x 2 + 5 = 17 x
b 9x 2 − 39x − 30 = 0
Answer
6x 2 + 5 = 17 x
a
6x 2 − 17 x + 5 = 0
(2 x − 5)(3x − 1) = 0
2 x − 5 = 0 or 3x − 1 = 0
x=
b
5
2
or
x=
Write in the form ax 2 + bx + c = 0.
Factorise.
Use the fact that if pq = 0 then p = 0 or q = 0.
Solve.
1
3
9x 2 − 39x − 30 = 0
Divide both sides by the common factor of 3.
3x 2 − 13x − 10 = 0
Factorise.
(3x + 2)( x − 5) = 0
3x + 2 = 0 or x − 5 = 0
x=−
Solve.
2
or x = 5
3
Original material © Cambridge University Press 2017
TIP
Divide by a common
factor first, if possible.
3
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
WORKED EXAMPLE 1.2
Solve
21
2
−
=1
2x x + 3
Answer
21
2
−
=1
2x x + 3
21( x + 3) − 4x = 2 x( x + 3)
2 x 2 − 11x − 63 = 0
7
or x = 9
2
WORKED EXAMPLE 1.3
Solve
3x 2 + 26 x + 35
=0
x2 + 8
Solve.
SA
M
4
PL
x=−
Expand brackets and rearrange.
Factorise.
(2 x + 7)( x − 9) = 0
2 x + 7 = 0 or x − 9 = 0
E
Multiply both sides by 2 x( x + 3).
Answer
3x 2 + 26 x + 35
=0
x2 + 8
Multiply both sides by x 2 + 8.
3x 2 + 26 x + 35 = 0
Factorise.
(3x + 5)( x + 7) = 0
3x + 5 = 0 or x + 7 = 0
x=−
Solve.
5
or x = −7
3
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
WORKED EXAMPLE 1.4
A rectangle has sides of length x cm and (6x − 7) cm.
x
The area of the rectangle is 90 cm 2.
Find the lengths of the sides of the rectangle.
6x – 7
Answer
Area = x(6x − 7) = 6x 2 − 7 x = 90
E
Rearrange.
6x 2 − 7 x − 90 = 0
Factorise
(2 x − 9)(3x + 10) = 0
2 x − 9 = 0 or 3x + 10 = 0
9
2
PL
x=
Solve.
x=−
or
10
3
Length is a positive quantity, so x = 4 21 .
When x = 4 21 , 6x − 7 = 20
The rectangle has sides of length 4 21 cm and 20 cm.
SA
M
EXPLORE 1.2
A
4(2 x
2
+ x − 6)
=1
B
( x 2 − 3x + 1)6 = 1
C
( x 2 − 3x + 1)(2 x
2
+ x − 6)
=1
1 Discuss with your classmates how you would solve each of these equations.
2 Solve
a equation A
b equation B
c equation C
Remember to check
each of your answers.
3 State how many values of x satisfy:
a equation A
b equation B
TIP
c equation C
4 Discuss your results.
Original material © Cambridge University Press 2017
5
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXERCISE 1A
1 Solve by factorisation.
a x 2 + 3x − 10 = 0
b x 2 − 7 x + 12 = 0
c
x 2 − 6x − 16 = 0
d 5x 2 + 19x + 12 = 0
e 20 − 7 x = 6x 2
f
x(10 x − 13) = 3
6
=0
x−5
a x−
b
2
3
+
=1
x x+2
E
2 Solve.
5x + 1 2 x − 1
−
= x2
4
2
d
5
3x
+
=2
x+3 x+4
e
3
1
+
=2
x + 1 x( x + 1)
f
3
1
1
+
=
x + 2 x − 1 ( x + 1)( x + 2)
3 Solve.
3x 2 + x − 10
=0
a
x2 − 7x + 6
d
x2 − 2x − 8
=0
x 2 + 7 x + 10
PL
c
b
x2 + x − 6
=0
x2 + 5
c
x2 − 9
=0
7 x + 10
e
6x 2 + x − 2
=0
x2 + 7x + 4
f
2 x 2 + 9x − 5
=0
x4 + 1
c
2( x
f
( x 2 − 7 x + 11)8 = 1
4 Find the real solutions of the following equations.
a 8( x
6
2
d 3(2 x
+ 2 x −15)
=1
2
=
+ 9 x + 2)
1
9
b 4(2 x
2
−11x +15)
=1
e ( x 2 + 2 x − 14)5 = 1
SA
M
5 The diagram shows a right-angled triangle with
sides 2 x cm, (2 x + 1) cm and 29 cm.
2
− 4 x + 6)
=8
TIP
2x
29
Check that your
answers satisfy the
original equation.
2
a Show that 2 x + x − 210 = 0.
b Find the lengths of the sides of the triangle.
2x + 1
6 The area of the trapezium is 35.75 cm 2.
Find the value of x.
PS
2
7 Solve ( x − 11x + 29)
(6 x 2 + x − 2)
WEB LINK
x+3
x–1
x
= 1.
1.2 Completing the square
Another method we can use for solving quadratic equations is completing the square.
The method of completing the square aims to rewrite a quadratic expression using only
one occurrence of the variable, making it an easier expression to work with.
If we expand the expressions ( x + d )2 and ( x − d )2 we obtain the results:
( x + d )2 = x 2 + 2 dx + d 2 and ( x − d )2 = x 2 − 2 dx + d 2
Original material © Cambridge University Press 2017
Try the Factorisable
quadratics resource
on the Underground
Mathematics website.
Chapter 1: Quadratics
Rearranging these gives the following important results:
KEY POINT 1.1
x 2 + 2 dx = ( x + d )2 − d 2 and x 2 − 2 dx = ( x − d )2 − d 2
To complete the square for x 2 + 10 x we can use the first of the above results as follows:
E
10 ÷ 2 = 5
x 2 + 10 x = ( x + 5)2 − 52
x 2 + 10 x = ( x + 5)2 − 25
PL
To complete the square for x 2 + 8x − 7, we again use the first result applied to the x 2 + 8x
part, as follows:
8÷2 = 4
x 2 + 8x − 7 = ( x + 4)2 − 42 − 7
x 2 + 8x − 7 = ( x + 4)2 − 23
To complete the square for 2 x 2 − 12 x + 5 we must first take a factor of 2 out of the first
two terms, so:
2 x 2 − 12 x + 5 = 2( x 2 − 6x ) + 5
SA
M
6÷2 = 3
x 2 − 6x = ( x − 3)2 − 32 , giving
2 x 2 − 12 x + 5 = 2 [( x − 3)2 − 9] + 5 = 2( x − 3)2 − 13
We can also use an algebraic method for completing the square as shown in Worked
example 1.5.
WORKED EXAMPLE 1.5
Express 2 x 2 − 12 x + 3 in the form p( x − q )2 + r, where p, q and r are constants to be found.
Answer
2 x 2 − 12 x + 3 = p( x − q )2 + r
Expand the brackets and simplify.
2 x 2 − 12 x + 3 = px 2 − 2 pqx + pq 2 + r
Compare coefficients of x 2 , coefficients of x
and the constant
2= p
(1)
−12 = −2 pq
(2)
3 = pq 2 + r
Substituting p = 2 in equation (2) gives q = 3
(3)
2 x 2 − 12 x + 3 = 2( x − 3)2 − 15
Substituting p = 2 and q = 3 in equation (3)
therefore gives r = −15
Original material © Cambridge University Press 2017
7
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
WORKED EXAMPLE 1.6
Express 4x 2 + 20 x + 5 in the form ( ax + b )2 + c, where a, b and c are constants to be found.
Answer
4x 2 + 20 x + 5 = ( ax + b )2 + c
4x 2 + 20 x + 5 = a 2 x 2 + 2 abx + b2 + c
E
Expanding the brackets and simplifying gives:
Comparing coefficients of x 2, coefficients of x and the constant gives
4 = a 2 (1)
20 = 2ab (2)
5 = b2 + c (3)
Equation (1) gives a = ±2
PL
Substituting a = 2 in equation (2) gives b = 5
Substituting b = 5 in equation (3) gives c = −20
4x 2 + 20 x + 5 = (2 x + 5)2 − 20
Alternatively:
Substituting a = −2 in equation (2) gives b = −5
Substituting b = −5 in equation (3) gives c = −20
4x 2 + 20 x + 5 = ( −2 x − 5)2 − 20 = (2 x + 5)2 − 20
8
SA
M
WORKED EXAMPLE 1.7
Use completing the square to solve the equation
Leave your answers in surd form.
5
3
+
= 1.
x+2 x−5
Answer
5
3
+
=1
x+2 x−5
5( x − 5) + 3( x + 2) = ( x + 2)( x − 5)
x 2 − 11x + 9 = 0
2
Multiply both sides by ( x + 2)( x − 5).
Expand brackets and collect terms.
Complete the square.
2
 x − 11  −  11  + 9 = 0

 2 
2 
2
 x − 11  = 85

2 
4
x−
11
=±
2
85
4
x =
11
85
±
2
2
x =
1
(11 ± 85 )
2
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
EXERCISE 1B
1 Express each of the following in the form ( x + a )2 + b.
a x 2 − 6x
e
x 2 + 4x + 8
b x 2 + 8x
f
c
x 2 − 4x − 8
x 2 − 3x
d x 2 + 15x
g x2 + 7x + 1
h x 2 − 3x + 4
2 Express each of the following in the form a ( x + b )2 + c.
b 3x 2 − 12 x − 1
c
2 x 2 + 5x − 1
3 Express each of the following in the form a − ( x + b )2 .
a 4x − x 2
b 8x − x 2
c
4 − 3x − x 2
d 9 + 5x − x 2
PL
4 Express each of the following in the form p − q ( x + r )2 .
a 7 − 8x − 2 x 2
d 2x2 + 7x + 5
E
a 2 x 2 − 12 x + 19
b 3 − 12 x − 2 x 2
c 13 + 4x − 2 x 2
d 2 + 5x − 3x 2
5 Solve by completing the square.
a x 2 + 8x − 9 = 0
d x 2 − 9x + 14 = 0
b x 2 + 4x − 12 = 0
c
x 2 − 2 x − 35 = 0
e x 2 + 3x − 18 = 0
f
x 2 + 9x − 10 = 0
6 Solve by completing the square. Leave your answers in surd form.
a x 2 + 4x − 7 = 0
d 2 x 2 − 4x − 5 = 0
b x 2 − 10 x + 2 = 0
c
x 2 + 8x − 1 = 0
e 2 x 2 + 6x + 3 = 0
f
2 x 2 − 8x − 3 = 0
5
3
+
= 2. Leave your answers in surd form.
x+2 x−4
SA
M
7 Solve
PS
8 The diagram shows a right-angled triangle with
sides x m, (2 x + 5) m and 10 m.
10
x
Find the value of x. Leave your answer in surd form.
2x + 5
PS
9 Find the real solutions of the equation (3x 2 + 5x − 7)4 = 1.
PS
49x 2
10 The path of a projectile is given by the equation y = ( 3 )x −
where x and
9000
y are measured in metres.
y
TIP
(x, y)
O
x
Range
a Find the range of this projectile.
b Find the maximum height reached by this projectile.
Original material © Cambridge University Press 2017
You will learn how
to derive formulae
such as this if you go
on to study Further
Mathematics
9
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
1.3 The quadratic formula
We can solve quadratic equations using the quadratic formula.
If ax 2 + bx + c = 0, where a, b and c are constants and a ≠ 0, then
KEY POINT 1.2
−b ± b 2 − 4ac
2a
E
x=
The quadratic formula can be proved by completing the square for the equation
ax 2 + bx + c = 0:
2
2
Divide both sides by a.
PL
2
ax 2 + bxax+ c+ =bx0 + c = 0
c
b 2 cb
x 2 + xx ++ a=x0+ a = 0
a
a
Complete the square.
2
2
 x + b  x +− b b − + cb = 0+ c = 0


2a2 a   2aa 
a

2a 
2
2
2
2
 x + b  x += bb  −= c b − c
24aa2 a4a 2 a

2 a 
2
2
2
2
 x + b  x += bb − 4=acb − 4ac
2
2 a 4a
4a 2

2 a 
2
b x + b b2= −±4acb − 4ac
x+
= ± 2a
2a
2a
2a
10
Rearrange the equation.
Write the right-hand side as a single fraction.
Square root both sides.
Subtract
b
from both sides.
2a
SA
M
2
bb2 − 4acb − 4ac
b
x = − x =± − 2 a ±
2 a Write the right-hand side as a single fraction.
2a
2a
x=
2
−b x± = b−2 b−±4acb − 4ac
2a
2a
WORKED EXAMPLE 1.8
Solve the equation 6x 2 − 3x − 2 = 0.
Write your answers correct to 3 significant figures.
Answer
Using a = 6, b = −3 and c = −2 in the quadratic formula gives:
x=
−( −3) ± ( −3)2 − 4 × 6 × ( −2)
2×6
x=
3 + 57
3 − 57
or x =
12
12
x = 0.879
or x = −0.379 (to 3 significant figures)
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
EXERCISE 1C
1 Solve using the quadratic formula. Give your answer correct to 2 decimal places.
a x 2 − 10 x − 3 = 0
b x 2 + 6x + 4 = 0
c
x 2 + 3x − 5 = 0
d 2 x 2 + 5x − 6 = 0
e 4x 2 + 7 x + 2 = 0
f
5x 2 + 7 x − 2 = 0
The area of the rectangle is 63 cm 2.
Find the value of x correct to 3 significant figures.
3 Rectangle A has sides of length x cm and (2 x − 4) cm.
E
2 A rectangle has sides of length x cm and (3x − 2) cm.
PL
Rectangle B has sides of length ( x + 1) cm and (5 − x ) cm.
Rectangle A and rectangle B have the same area.
Find the value of x correct to 3 significant figures.
5
2
+
= 1.
x −3 x +1
Give your answers correct to 3 significant figures.
4 Solve the equation
2
5 Solve the quadratic equation ax − bx + c = 0 giving your answers in terms
of a, b and c.
SA
M
How do the solutions of this equation relate to the solutions of
the equation ax 2 + bx + c = 0?
1.4 Solving simultaneous equations (one linear and one quadratic)
In this section, we shall learn how to solve simultaneous equations where one equation is
linear and the second equation is quadratic.
y
y = x2 – 4
y = 2x – 1
(3, 5)
O
x
(–1, –3)
The diagram shows the graphs of y = x 2 − 4 and y = 2 x − 1.
The coordinates of the points of intersection of the two graphs are ( −1, −3) and (3, 5).
It follows that x = −1, y = −3 and x = 3, y = 5 are the solutions of the simultaneous
equations y = x 2 − 4 and y = 2 x − 1.
Original material © Cambridge University Press 2017
WEB LINK
Try the Quadratic
solving sorter resource
on the Underground
Mathematics website.
11
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
The solutions can also be found algebraically:
y = x2 − 4
y = 2x − 1
(1)
(2)
Substitute for y from equation (2) into equation (1):
x2 − 4
0
0
3
Rearrange
Factorise
Substituting x = −1 into equation (2) gives y = −2 − 1 = −3
Substituting x = 3 into equation (2) gives y = 6 − 1 = 5
PL
The solutions are: x = −1, y = −3 and x = 3, y = 5
E
2x − 1 =
x2 − 2x − 3 =
( x + 1)( x − 3) =
x = −1 or x =
In general, an equation in x and y is called quadratic if it has the form
ax 2 + bxy + cy2 + dx + ey + f = 0, where at least one of a, b and c is non-zero.
Our technique for solving one linear and one quadratic equation will work for these more general
quadratics too. (The graph of a general quadratic function such as this is called a conic.)
WORKED EXAMPLE 1.9
Solve the simultaneous equations.
2x + 2 y = 7
x 2 − 4 y2 = 8
SA
M
12
Answer
2x + 2 y = 7
(1)
2
(2)
2
x − 4y = 8
7 − 2y
2
Substitute for x in equation (2):
From equation (1), x =
2
 7 − 2 y  − 4 y2 = 8
 2 
49 − 28 y + 4 y2
− 4 y2 = 8
4
49 − 28 y + 4 y2 − 16 y2 = 32
12 y2 + 28 y − 17 = 0
Expand brackets.
Multiply both sides by 4.
Rearrange.
Factorise.
(6 y + 17)(2 y − 1) = 0
y=−
17
1
or y =
6
2
Substituting y = −
17
19
in equation (1) gives x =
6
3
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
1
in equation (1) gives x = 3
2
19
17
1
The solutions are: x = , y = −
and x = 3, y =
3
6
2
Substituting y =
Alternative method:
From equation (1), 2 y = 7 − 2 x
E
Substitute for 2 y in equation (2):
x 2 − (7 − 2 x )2 = 8
Expand brackets.
x 2 − 49 + 28x − 4x 2 = 8
(3x − 19)( x − 3) = 0
x=
19
or x = 3
3
The solutions are: x =
Factorise.
PL
3x 2 − 28x + 57 = 0
Rearrange.
19
17
1
,y=−
and x = 3, y =
3
6
2
EXERCISE 1D
SA
M
1 Solve the simultaneous equations.
a
y = 6−x
b x + 4y = 6
y = x2
x 2 + 2 xy = 8
d y = 3x − 1
e
8x 2 − 2 xy = 4
g 2x + y = 8
h 2y − x = 5
f
k x − 4y = 2
2
x − 5xy + y = 1
m 2 x + 3 y + 19 = 0
xy = 12
n x + 2y = 5
2x2 + 3 y = 5
x 2 + y2 = 10
4x − 3 y = 5
x 2 + 3xy = 10
i
2 x 2 − 3 y2 = 15
5x − 2 y = 23
2
x − 2y = 6
x 2 + y2 = 100
x 2 − 4xy = 20
xy = 8
j
c 3 y = x + 10
x + 2y = 6
x 2 + y2 + 4xy = 24
l
2 x − y = 14
y 2 = 8x + 4
o x − 12 y = 30
2 y2 − xy = 20
2 The sum of two numbers is 26. The product of the two numbers is 153.
a What are the two numbers?
b If instead the product is 150 (and the sum is still 26), what would the two
numbers now be?
3 The perimeter of a rectangle is 15.8 cm and its area is 13.5 cm 2. Find the lengths
of the sides of the rectangle.
Original material © Cambridge University Press 2017
13
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
4 The sum of the perimeters of two squares is 50 cm and the sum of the areas is
93.25 cm 2.
Find the side length of each square.
5 The sum of the circumferences of two circles is 36 π cm and the sum of the
areas is 170 π cm 2.
E
Find the radius of each circle.
6 A cuboid has sides of length 5 cm, x cm and y cm. Given that x + y = 20.5 and
the volume of the cuboid is 360 cm 3, find the value of x and the value of y.
TIP
18 cm
PL
7 The diagram shows a solid formed by joining a
hemisphere of radius r cm to a cylinder of radius r cm
and height h cm.
The total height of the solid is 18 cm and the surface
area is 205 π cm 2.
h
Find the value of r and the value of h.
r
The surface area, A,
of a sphere with radius
r is A = 4 π r 2.
8 The line y = 2 − x cuts the curve 5x 2 − y2 = 20 at the points A and B.
a Find the coordinates of the points A and B.
b Find the length of the line AB.
9 The line 2 x + 5 y = 1 meets the curve x 2 + 5xy − 4 y2 + 10 = 0 at the points A and B.
14
SA
M
a Find the coordinates of the points A and B.
b Find the mid-point of the line AB.
10 The line 7 x + 2 y = −20 intersects the curve x 2 + y2 + 4x + 6 y − 40 = 0 at the
points A and B. Find the length of the line AB.
11 The line 7 y − x = 25 cuts the curve x 2 + y2 = 25 at the points A and B.
Find the equation of the perpendicular bisector of the line AB.
12 The straight line y = x + 1 intersects the curve x 2 − y = 5 at the points A and B.
Given that A lies below the x axis and the point P lies on AB such that
AP : PB = 4 : 1, find the coordinates of P.
13 The line x − 2 y = 1 intersects the curve x + y2 = 9 at two points A and B.
Find the equation of the perpendicular bisector of the line AB.
PS
14 aSplit 10 into two parts so that the difference between the squares of the parts
is 60.
b Split N into two parts so that the difference between the squares of the parts
is D.
Original material © Cambridge University Press 2017
WEB LINK
Try the Elliptical
crossings resource
on the Underground
Mathematics website.
Chapter 1: Quadratics
1.5 Solving more complex quadratic equations
You may be asked to solve an equation that is quadratic in some function of x.
WORKED EXAMPLE 1.10
Answer
Method 1: Substitution method
4x 4 − 37 x 2 + 9 = 0
Let y = x 2 .
4 y2 − 37 y + 9 = 0
y=
1
or y = 9
4
Substitute x 2 for y.
PL
(4 y − 1)( y − 9) = 0
E
Solve the equation 4x 4 − 37 x 2 + 9 = 0.
1
or x 2 = 9
4
1
x=±
or x = ±3
2
Method 2: Factorise directly
x2 =
4x 4 − 37 x 2 + 9 = 0
2
2
SA
M
(4x − 1)( x − 9) = 0
x2 =
x=±
1
or x 2 = 9
4
1
or x = ±3
2
WORKED EXAMPLE 1.11
Solve the equation x − 4 x − 12 = 0.
Answer
x − 4 x − 12 = 0
Let y =
x.
y2 − 4 y − 12 = 0
( y − 6)( y + 2) = 0
y = 6 or y = −2
x = 6 or
x = −2
Substitute
x for y.
x = −2 has no solutions as x is never negative.
∴ x = 36
Original material © Cambridge University Press 2017
15
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
WORKED EXAMPLE 1.12
Solve the equation 3(9x ) − 28(3x ) + 9 = 0.
Answer
3(3x )2 − 28(3x ) + 9 = 0
Let y = 3x.
E
3 y2 − 28 y + 9 = 0
(3 y − 1)( y − 9) = 0
3x =
1
or y = 9
3
1
or 3x = 9
3
x = −1 or x = 2
EXERCISE 1E
Substitute 3x for y.
1
= 3−1 and 9 = 32.
3
PL
y=
1 Find the real values of x satisfying the following equations.
a x 4 − 13x 2 + 36 = 0
16
c
x 4 − 6x 2 + 5 = 0
e 3x 4 + x 2 − 4 = 0
f
8x 6 − 9x 3 + 1 = 0
SA
M
d 2 x 4 − 11x 2 + 5 = 0
b x6 − 7x3 − 8 = 0
g
x 4 + 2 x 2 − 15 = 0
h x 4 + 9x 2 + 14 = 0
i
x8 − 15x 4 − 16 = 0
j
32 x10 − 31x5 − 1 = 0
k
9
5
+
=4
x 4 x2
l
8
7
+
=1
x6 x3
c
6x − 17 x + 5 = 0
f
3 x+
2 Solve.
a 2 x − 9 x + 10 = 0
b
x ( x + 1) = 6
d 10 x + x − 2 = 0
e 8x + 5 = 14 x
5
= 16
x
3 The curve y = 2 x and the line 3 y = x + 8 intersect at the points A and B.
a Write down an equation satisfied by the x-coordinates of A and B.
b Solve your equation in part a and hence find the coordinates of A and B.
c
PS
Find the length of the line AB.
4 The graph shows y = ax + b x + c for x ù 0. The graph crosses the x-axis
49
at the points ( 1, 0 ) and  , 0  and it meets the y-axis at the point (0, 7).
 4 
Find the value of a, the value of b and the value of c.
y
7
O
Original material © Cambridge University Press 2017
1
49
4
x
Chapter 1: Quadratics
PS
5 The graph shows y = a (22 x ) + b(2 x ) + c.
The graph crosses the axes at the points (2, 0), (4, 0) and (0, 90).
Find the value of a, the value of b and the value of c.
y
90
2
4
x
E
O
1.6 Maximum and minimum values of a quadratic function
The general form of a quadratic function is f( x ) = ax 2 + bx + c, where a, b and c are
constants and a ≠ 0.
PL
The shape of the graph of the function f( x ) = ax 2 + bx + c is called a parabola.
The orientation of the parabola depends on the value of a, the coefficient of x 2.
TIP
A point where the
gradient is zero is
called a stationary
point or a turning point.
If a , 0, the curve has a maximum
point that occurs at the highest point
of the curve.
SA
M
If a . 0, the curve has a minimum
point that occurs at the lowest point
of the curve.
In the case of a parabola, we also call this point the vertex of the parabola.
Every parabola has a line of symmetry that passes through the vertex.
One important skill that we will develop during this course is ‘graph sketching’.
A sketch graph needs to show the key features and behaviour of a function.
When we sketch the graph of a quadratic function, the key features are:
●●
●●
●●
the general shape of the graph
the axis intercepts
the coordinates of the vertex.
Depending on the context we should show some or all of these.
The skills you developed earlier in this chapter should enable you to draw a clear sketch
graph for any quadratic function.
Original material © Cambridge University Press 2017
WEB LINK
Try the Quadratic
symmetry resource
on the Underground
Mathematics
website for a further
explanation of this.
17
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
DID YOU KNOW?
E
If we rotate a parabola about its axis of symmetry, we
obtain a three-dimensional shape called a paraboloid.
Satellite dishes are paraboloid shapes. They have the
special property that light rays are reflected to meet at a
single point, if they are parallel to the axis of symmetry
of the dish. This single point is called the focus of the
satellite dish. A receiver at the focus of the paraboloid
then picks up all the information entering the dish.
f( x ) = x 2 − 3x − 4
PL
WORKED EXAMPLE 1.13
a Find the axis crossing points for the graph of y = f( x ).
b Sketch the graph of y = f( x ) and find the coordinates of the vertex.
Answer
a y = x 2 − 3x − 4
When x = 0, y = −4.
When y = 0, x 2 − 3x − 4 = 0
( x + 1)( x − 4) = 0
x = −1 or x = 4
SA
M
18
Axes crossing points are: (0, −4), ( −1, 0) and (4, 0).
b The line of symmetry cuts the x-axis midway between the axis intercepts of −1
and 4.
x=
y
–1
3
2
y = x 2 – 3x – 4
O
4
x
–4
(
( 32 , –254
Hence the line of symmetry is x =
2
3
2
3
3
3
, y =   −3  − 4
 2
 2
2
25
y=−
4
Since a . 0, the curve is U-shaped.
3
25 
Minimum point =  , −
2
4 
When x =
Original material © Cambridge University Press 2017
TIP
Write your answer in
fraction form.
Chapter 1: Quadratics
Completing the square is an alternative method that can be used to help sketch the graph
of a quadratic function.
Completing the square for x 2 − 3x − 4 gives:
2
2
3
3
x 2 − 3x − 4 =  x −  −   − 4

 2
2
This part of the expression is a square so it will be
at least zero. The smallest value it can be is 0. This
3
occurs when x = ..
2
2
E
3
25
=x−  −


2
4
2
KEY POINT 1.3
PL
25
3
3
25
and this minimum occurs when x = .
is −
The minimum value of  x −  −


4
2
2
4
3
25 

2
So the function f( x ) = x − 3x − 4 has a minimum point at
,−
.
2
4 
3
The line of symmetry is x = .
2
If f( x ) = ax 2 + bx + c is written in the form f( x ) = a( x − h )2 + k, then
b
●● the line of symmetry is x = h = −
2a
●● if a . 0, there is a minimum point at ( h, k )
●●
if a , 0, there is a maximum point at ( h, k ).
SA
M
WORKED EXAMPLE 1.14
Sketch the graph of y = 16x − 7 − 4x 2 .
Answer
Completing the square gives:
16x − 7 − 4x 2 = 9 − 4( x − 2)2
The maximum value of 9 − 4( x − 2)2 is 9 and this
maximum occurs when x = 2.
This part of the expression is a square so ( x − 2)2 ù 0.
The smallest value it can be is 0. This occurs when x = 2.
Since this is being subtracted from 9, the whole expression
is greatest when x = 2.
So the function f( x ) = 16x − 7 − 4x 2 has
a maximum point at (2, 9).
The line of symmetry is x = 2.
y
(2, 9)
When x = 0, y = −7
When y = 0, 9 − 4( x − 2)2 = 0
9
( x − 2)2 =
4
x−2 = ±
x = 3 21 or x =
3
2
1
2
1
2
y = 16x – 7 – 4x2
31
2
–7
x=2
Original material © Cambridge University Press 2017
x
19
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXERCISE 1F
1 Use the symmetry of each quadratic function to find the maximum or minimum points.
Sketch each graph, showing all axis crossing points.
a
y = x 2 − 6x + 8
b y = x 2 + 5x − 14
c
y = 2 x 2 + 7 x − 15
d y = 12 + x − x 2
E
2 a Express 2 x 2 − 8x + 5 in the form a ( x + b )2 + c, where a, b and c are integers.
b Write down the equation of the line of symmetry for the graph of y = 2 x 2 − 8x + 1.
3 a Express 7 + 5x − x 2 in the form a − ( x + b )2, where a, and b are constants.
PL
b Find the coordinates of the turning point of the curve y = 7 + 5x − x 2, stating whether it is a maximum or
a minimum point.
4 aExpress 2 x 2 + 9x + 4 in the form a ( x + b )2 + c, where a, b and c are constants.
b Write down the coordinates of the vertex of the curve y = 2 x 2 + 9x + 4, and state whether this is a
maximum or a minimum point.
5 Find the minimum value of x 2 − 7 x + 8 and the corresponding value of x.
6 a Write 1 + x − 2 x 2 in the form p − 2( x − q )2.
b Sketch the graph of y = 1 + x − 2 x 2 .
20
7 Prove that the graph of y = 4x 2 + 2 x + 5 does not intersect the x-axis.
8 Find the equations of parabolas A, B and C.
SA
M
PS
y
12
A
10
B
8
6
4
2
–4
–2
O
–2
–4
2
4
8 x
6
C
–6
–8
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
PS
9 The diagram shows eight parabolas.
The equations of two of the parabolas are y = x 2 − 6x + 13 and y = − x 2 − 6x − 5.
a Identify these two parabolas and find the equation of each of the other parabolas.
b Use graphing software to create your own parabola pattern.
y
A
G
x
PL
O
D
H
B
E
F
E
C
(This question is an adaptation of Which parabola? on the Underground Mathematics website
and was developed from an original idea from NRICH.)
PS
10 A parabola passes through the points (0, −24), ( −2, 0) and (4, 0).
Find the equation of the parabola.
11 A parabola passes through the points ( −2, −3), (2, 9) and (6, 5).
SA
M
PS
Find the equation of the parabola.
P
12 Prove that any quadratic that has its vertex at ( p, q ) has an equation of the form y = ax 2 − 2 apx + ap2 + q
for some non-zero real number a.
1.7 Solving quadratic inequalities
We already know how to solve linear inequalities.
The following text shows two examples.
Solve 2( x + 7) , − 4
2 x + 14 , − 4
2 x , −18
Expand brackets.
Subtract 14 from both sides.
Divide both sides by 2.
x , −9
Solve 11 − 2 x ù 5
−2 x ù −6
Subtract 11 from both sides.
Divide both sides by −2.
xø3
Original material © Cambridge University Press 2017
21
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
The second of the previous examples uses the important rule that:
KEY POINT 1.4
If we multiply or divide both sides of an inequality by a negative number then the inequality sign
must be reversed.
WORKED EXAMPLE 1.15
Solve x 2 − 5x − 14 . 0.
PL
y
E
Quadratic inequalities can be solved by sketching a graph and considering when the graph
is above or below the x-axis.
y = x2 – 5x – 14
Answer
Sketch the graph of y = x 2 − 5x − 14.
When y = 0, x 2 − 5x − 14 = 0
( x + 2)( x − 7) = 0
x = −2 or x = 7
TIP
+
+
–2
O
7
For the sketch graph,
you only need to
identify which way up
the graph is and where
the x-intercepts are:
you do not need to find
the vertex or the
y-intercept.
x
So the x-axis crossing points are −2 and 7.
The solution is x , −2 or x . 7.
SA
M
22
For x 2 − 5x − 14 . 0 we need to find the
range of values of x for which the curve is
positive (above the x-axis).
–
WORKED EXAMPLE 1.16
Solve 2 x 2 + 3x ø 27.
Answer
Rearranging: 2 x 2 + 3x − 27 ø 0
Sketch the graph of y = 2 x 2 + 3x − 27.
When y = 0, 2 x 2 + 3x − 27 = 0
(2 x + 9)( x − 3) = 0
y
y = 2x2 + 3x – 27
+
+
O
–4 1
2
x = −4 21 or x = 3
So the x-axis intercepts are −4 21 and 3.
For 2 x 2 + 3x − 27 ø 0 we need to find the range of values of x
for which the curve is either zero or negative (below the x-axis).
The solution is − 4 21 < x < 3.
–
Original material © Cambridge University Press 2017
3
x
Chapter 1: Quadratics
EXPLORE 1.3
Ivan is asked to solve the inequality
2x − 4
ù 7.
x
This is his solution:
Subtract 2x from both sides:
−4 ù 5x
x ø−
Divide both sides by 5:
4
5
Anika checks to see if x = −1 satisfies the original inequality.
PL
She writes:
E
2x − 4 ù 7x
Multiply both sides by x:
When x = −1: (2(−1) − 4) ÷ (−1) = 6
Hence x = −1is a value of x that does not satisfy the original inequality.
So Ivan’s solution must be incorrect!
Discuss Ivan’s solution with your classmates and explain Ivan’s error.
How could Ivan have approached this problem to obtain a correct solution?
SA
M
EXERCISE 1G
1 Solve.
a x( x − 3) ø 0
b ( x − 3)( x + 2) . 0
c
( x − 6)( x − 4) ø 0
d (2 x + 3)( x − 2) , 0
e (5 − x )( x + 6) ù 0
f
(1 − 3x )(2 x + 1) , 0
a x 2 − 25 ù 0
b x 2 + 7 x + 10 ø 0
c
x 2 + 6x − 7 . 0
d 14x 2 + 17 x − 6 ø 0
e 6x 2 − 23x + 20 , 0
f
4 − 7x − 2x2 , 0
b 15x , x 2 + 56
c
x( x + 10) ø 12 − x
d x + 4x , 3( x + 2)
e ( x + 3)(1 − x ) , x − 1
f
(4x + 3)(3x − 1) , 2 x( x + 3)
g ( x + 4)2 ù 25
h ( x − 2)2 . 14 − x
i
6x( x + 1) , 5(7 − x )
2 Solve.
3 Solve.
a x 2 , 36 − 5x
2
4 Find the range of values of x for which
5
, 0.
2 x 2 + x − 15
5 Find the set of values of x for which:
a x 2 − 3x ù 10
and ( x − 5)2 , 4
b x 2 + 4x − 21 ø 0 and x 2 − 9x + 8 . 0
c
x2 + x − 2 . 0
and x 2 − 2 x − 3 ù 0
6 Find the range of values of x for which 2 x
2
− 3x − 40
. 1.
Original material © Cambridge University Press 2017
23
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
7 Solve.
x
ù3
a
x −1
d
x 2 − 2 x − 15
ù0
x−2
b
x( x − 1)
.x
x +1
c
x2 − 9
ù4
x −1
e
x 2 + 4x − 5
ø0
x2 − 4
f
x−3 x+2
ù
x+4 x−5
1.8 The number of roots of a quadratic equation
E
If f( x ) is a function, then we call the solutions to the equation f( x ) = 0 the roots of f( x ).
Consider solving the following three quadratic equations of the form ax 2 + bx + c = 0
−b ± b2 − 4ac
using the formula x =
.
2a
x=
x 2 + 6x + 9 = 0
x 2 + 2x + 6 = 0
−6 ± 62 − 4 × 1 × 9
2 ×1
−6 ± 0
x=
2
x = −3 or x = −3
−2 ± 2 2 − 4 × 1 × 6
2 ×1
−2 ± −20
x=
2
no real solution
two equal real roots
no real roots
−2 ± 2 2 − 4 × 1 × ( −8 )
x=
2 ×1
−2 ± 36
x=
2
x = 2 or x = −4
two distinct real roots
PL
x 2 + 2x − 8 = 0
x=
The part of the quadratic formula underneath the square root sign is called the discriminant.
KEY POINT 1.5
The discriminant of ax 2 + bx + c = 0 is b 2 − 4ac.
SA
M
24
The sign (positive, zero or negative) of the discriminant tells us how many roots there are
for a particular quadratic equation.
b 2 − 4 ac
.0
=0
,0
Nature of roots
two distinct real roots
two equal real roots (or 1 repeated real root)
no real roots
There is a connection between the roots of the quadratic equation ax 2 + bx + c = 0 and the
corresponding curve y = ax 2 + bx + c.
b 2 − 4 ac
.0
Nature of roots of
ax 2 + bx + c = 0
Shape of curve y = ax 2 + bx + c
two distinct real roots
The curve cuts the x-axis at two distinct points.
.0
=0
,0
two equal real roots (or 1
repeated real root)
no real roots
or
,0
The curve touches the x-axis at one point.
.0
or
,0
The curve is entirely above or entirely below the x-axis.
x
a.0
or
a,0
x
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
WORKED EXAMPLE 1.17
Find the values of k for which the equation 4x 2 + kx + 1 = 0 has two equal roots.
Answer
b2 − 4ac = 0
For two equal roots:
k2 − 4 × 4 × 1 = 0
E
k 2 = 16
k = −4 or k = 4
PL
WORKED EXAMPLE 1.18
Find the values of k for which x 2 − 5x + 9 = k (5 − x ) has two equal roots.
Answer
x 2 − 5x + 9 = k (5 − x )
x 2 − 5x + 9 − 5 k + kx = 0
x 2 + ( k − 5)x + 9 − 5 k = 0
Rearrange the equation in to form ax 2 + bx + c = 0
For two equal roots: b2 − 4ac = 0
( k − 5)2 − 4 × 1 × (9 − 5 k ) = 0
2
k − 10 k + 25 − 36 + 20 k = 0
SA
M
k 2 + 10 k − 11 = 0
( k + 11)( k − 1) = 0
k = −11 or k = 1
WORKED EXAMPLE 1.19
Find the values of k for which kx 2 − 2 kx + 8 = 0 has two distinct roots.
Answer
kx 2 − 2 kx + 8 = 0
For two distinct roots:
b2 − 4ac . 0
( −2 k )2 − 4 × k × 8 . 0
+
+
2
4 k − 32 k . 0
4 k ( k − 8) . 0
0
–
8
Critical values are 0 and 8.
Note that the critical values are where 4 k ( k − 8) = 0 .
Hence k , 0 or k . 8
Original material © Cambridge University Press 2017
k
25
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXERCISE 1H
1 Find the discriminant for each equation and hence decide if the equation has two
distinct roots, two equal roots or no real roots.
a
x 2 − 12 x + 36 = 0
d 4x 2 − 4x + 1 = 0
b x 2 + 5x − 36 = 0
c
x 2 + 9x + 2 = 0
2x2 − 7x + 8 = 0
f
3x 2 + 10 x − 2 = 0
e
4
.
x
E
2 Use the discriminant to determine the nature of the roots of 2 − 5x =
3 The equation x 2 + bx + c = 0 has roots −5 and 7.
Find the value of b and the value of c.
4 Find the values of k for which the following equations have two equal roots.
x 2 + kx + 4 = 0
b 4x 2 + 4( k − 2)x + k = 0
c
( k + 2)x 2 + 4 k = (4 k + 2)x
d x 2 − 2 x + 1 = 2 k ( k − 2)
e
( k + 1)x 2 + kx − 2 k = 0
f
PL
a
4x 2 − ( k − 2)x + 9 = 0
5 Find the values of k for which the following equations have two distinct roots.
26
a
x 2 + 8x + 3 = k
c
kx 2 − 4x + 2 = 0
e
2 x 2 = 2( x − 1) + k
b 2 x 2 − 5x = 4 − k
d kx 2 + 2( k − 1)x + k = 0
f
kx 2 + (2 k − 5)x = 1 − k
6 Find the values of k for which the following equations have no real roots.
kx 2 − 4x + 8 = 0
b 3x 2 + 5x + k + 1 = 0
c
2 x 2 + 8x − 5 = kx 2
d 2 x 2 + k = 3( x − 2)
e
kx 2 + 2 kx = 4x − 6
f
SA
M
a
kx 2 + kx = 3x − 2
7 The equation kx 2 + px + 5 = 0 has repeated real roots.
Find k in terms of p.
8 Find the range of values of k for which the equation kx 2 − 5x + 2 = 0 has real
roots.
P
9 Prove that the roots of the equation 2 kx 2 + 5x − k = 0 are real and distinct for
all real values of k.
P
10 Prove that the roots of the equation x 2 + ( k − 2)x − 2 k = 0 are real and distinct
for all real values of k.
P
11 Prove that x 2 + kx + 2 = 0 has real roots if k ù 2 2.
For which other values of k does the equation have real roots?
Original material © Cambridge University Press 2017
WEB LINK
Try the Discriminating
resource on the
Underground
Mathematics website.
Chapter 1: Quadratics
1.9 Intersection of a line and a quadratic curve
When considering the intersection of a straight line and a parabola, there are three
possible situations.
Situation 2
Situation 3
E
Situation 1
one point of intersection
no points of intersection
The line cuts the curve at two
distinct points.
The line touches the curve at
one point. This means that the
line is a tangent to the curve.
The line does not intersect the
curve.
PL
two points of intersection
We have already learnt that to find the points of intersection of a straight line and a
quadratic curve, we solve their equations simultaneously.
The discriminant of the resulting equation then enables us to say how many points of
intersection there will be. The three possible situations are shown in the following table.
b 2 − 4 ac
Nature of roots
Line and curve
.0
two distinct real roots
two distinct points of intersection
=0
two equal real roots (repeated roots) one point of intersection (line is a tangent)
,0
no real roots
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no points of intersection
WORKED EXAMPLE 1.20
Find the value of k for which y = x + k is a tangent to the curve y = x 2 + 5x + 2.
Answer
x 2 + 5x + 2 = x + k
x 2 + 4x + (2 − k ) = 0
Since the line is a tangent to the curve, the discriminant of the quadratic must be zero, so:
b2 − 4ac = 0
42 − 4 × 1 × (2 − k )
16 − 8 + 4 k
4k
k
=
=
=
=
0
0
−8
−2
Original material © Cambridge University Press 2017
27
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
WORKED EXAMPLE 1.21
Find the set of values of k for which y = kx − 1 intersects the curve y = x 2 − 2 x at two distinct points.
Answer
x 2 − 2 x = kx − 1
x − ( k + 2)x + 1 = 0
2
b2 − 4ac . 0
( k + 2) − 4 × 1 × 1 . 0
2
+
k 2 + 4k . 0
k ( k + 4) . 0
Hence, k , −4 or k . 0
+
–4
k
0
PL
Critical values are −4 and 0.
E
Since the line intersects the curve at two distinct points, we must have discriminant . 0.
–
This next example involves a more general quadratic equation. Our techniques for finding
the conditions for intersection of a straight line and a quadratic equation will work for this
more general quadratic equation too.
WORKED EXAMPLE 1.22
Find the set of values of k for which the line 2 x + y = k does not intersect the curve xy = 8 .
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28
Answer
Substituting y = k − 2 x into xy = 8 gives:
x( k − 2 x ) = 8
2 x 2 − kx + 8 = 0
Since the line and curve do not intersect, we must have discriminant , 0.
b2 − 4ac , 0
( − k )2 − 4 × 2 × 8 , 0
k 2 − 64 , 0
( k + 8)( k − 8) , 0
Critical values are −8 and 8.
+
+
–8
8
–
Hence, −8 , k , 8
Original material © Cambridge University Press 2017
k
Chapter 1: Quadratics
EXERCISE 1I
1 Find the values of k for which the line y = kx + 1 is a tangent to the curve y = x 2 − 7 x + 2.
2 Find the values of k for which the x-axis is a tangent to the curve y = x 2 − ( k + 3)x + (3k + 4).
5
.
x−2
Can you explain graphically why there is only one such value of k? (You may want to use graph-drawing
software to help with this.)
E
3 Find the value of k for which the line x + ky = 12 is a tangent to the curve y =
4 The line y = k − 3x is a tangent to the curve x 2 + 2 xy − 20 = 0.
a Find the possible values of k.
PL
b For each of these values of k, find the coordinates of the point of contact of the tangent with the curve.
5 Find the values of m for which the line y = mx + 6 is a tangent to the curve y = x 2 − 4x + 7.
For each of these values of m, find the coordinates of the point where the line touches the curve.
6 Find the set of values of k for which the line y = 2 x − 1 intersects the curve y = x 2 + kx + 3 at two distinct
points.
7 Find the set of values of k for which the line x + 2 y = k intersects the curve xy = 6 at two distinct points.
8 Find the set of values of k for which the line y = k − x cuts the curve y = 5 − 3x − x 2 at two distinct points.
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9 Find the set of values of m for which the line y = mx + 5 does not meet the curve y = x 2 − x + 6.
10 Find the set of values of k for which the line y = 2 x − 10 does not meet the curve y = x 2 − 6x + k.
11 Find the value of k for which the line y = kx + 6 is a tangent to the curve x 2 + y2 − 10 x + 8 y = 84.
P
12 The line y = mx + c is a tangent to the curve y = x 2 − 4x + 4.
Prove that m2 + 8 m + 4c = 0.
P
13 The line y = mx + c is a tangent to the curve ax 2 + by2 = c, where a, b, c and m are constants.
abc − a
.
Prove that m2 =
b
Original material © Cambridge University Press 2017
29
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Checklist of learning and understanding
Quadratic equations can be solved by
●●
factorisation
●●
completing the square
●●
using the quadratic formula x =
−b ± b 2 − 4ac
.
2a
●●
Rearrange the linear equation to make either x or y the subject.
●●
Substitute this for x or y in the quadratic equation and then solve.
Maximum and minimum points and lines of symmetry
E
Solving simultaneous equations where one is linear and one is quadratic
PL
For a quadratic function f( x ) = ax 2 + bx + c that is written in the form f( x ) = a ( x − h )2 + k
b
●● the line of symmetry is x = h = −
.
2a
●●
if a . 0, there is a minimum point at ( h, k ).
●●
if a , 0, there is a maximum point at ( h, k ).
Quadratic equation ax 2 + bx + c = 0 and corresponding curve y = ax 2 + bx + c
Discriminant = b 2 − 4ac.
●●
If b 2 − 4ac . 0, then the equation ax 2 + bx + c = 0 has two distinct real roots.
●●
If b 2 − 4ac = 0, then the equation ax 2 + bx + c = 0 has two equal real roots.
●●
If b 2 − 4ac , 0, then the equation ax 2 + bx + c = 0 has no real roots.
●●
The condition for a quadratic equation to have real roots is b 2 − 4ac ù 0.
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30
●●
Intersection of a line and a general quadratic curve
●●
If a line and a general quadratic curve intersect at one point then the line is a tangent to the curve at that point.
●●
Solving simultaneously the equations for the line and the curve gives an equation of the form ax 2 + bx + c = 0.
●●
b 2 − 4ac gives information about the intersection of the line and the curve.
b 2 − 4 ac
Nature of roots
Line and parabola
.0
two distinct real roots
two distinct points of intersection
=0
two equal real roots
one point of intersection (line is a tangent)
,0
no real roots
no points of intersection
Original material © Cambridge University Press 2017
Chapter 1: Quadratics
END-OF-CHAPTER REVIEW EXERCISE 1
2
A curve has equation y = 2 xy + 5 and a line has equation 2 x + 5 y = 1.
The curve and the line intersect at the points A and B. Find the coordinates of the mid-point
of the line AB.
[4]
a Express 9x 2 − 15x in the form (3x − a )2 − b.
[2]
b Find the set of values of x that satisfy the inequality 9x 2 − 15x , 6. [2]
36
25
+ 4 = 2 .
x4
x
E
1
Find the real roots of the equation
4
Find the set of values of k for which the line y = kx − 3 intersects the curve y = x 2 − 9x at two
distinct points.
5
Find the set of values of the constant k for which the line y = 2 x + k meets the curve y = 1 + 2 kx − x 2
at two distinct points.
[5]
6
a Find the coordinates of the vertex of the parabola y = 4x 2 − 12 x + 7. [4]
b Find the values of the constant k for which the line y = kx + 3 is a tangent to the
curve y = 4x 2 − 12 x + 7. [3]
8
9
[4]
[4]
A curve has equation y = 5 − 2 x + x 2 and a line has equation y = 2 x + k, where k is a constant.
a
Show that the x-coordinates of the points of intersection of the curve and the line are
given by the equation x 2 − 4x + (5 − k ) = 0. [1]
b
For one value of k, the line intersects the curve at two distinct points A and B, where the
coordinates of A are ( −2, 13). Find the coordinates of B.
[3]
c
For the case where the line is a tangent to the curve at a point C , find the value of k and the
coordinates of C.
[4]
SA
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7
PL
3
A curve has equation y = x 2 − 5x + 7 and a line has equation y = 2 x − 3.
a
Show that the curve lies above the x-axis.
[3]
b
Find the coordinates of the points of intersection of the line and the curve.
[3]
c
Write down the set of values of x that satisfy the inequality x 2 − 5x + 7 , 2 x − 3. [1]
A curve has equation y = 10 x − x 2 .
a
Express 10 x − x 2 in the form a − ( x + b )2 .
[3]
b
Write down the coordinates of the vertex of the curve.
[2]
c
Find the set of values of x for which y ø 9. [3]
10 A line has equation y = kx + 6 and a curve has equation y = x 2 + 3x + 2 k, where k is a constant.
i
ii
For the case where k = 2, the line and the curve intersect at points A and B.
Find the distance AB and the coordinates of the mid-point of AB.
[5]
Find the two values of k for which the line is a tangent to the curve.
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2011
Original material © Cambridge University Press 2017
31
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
11 A curve has equation y = x 2 − 4x + 4 and a line has the equation y = mx, where m is a constant.
i
Find the coordinates of the mid-point of AB.
[4]
Find the non-zero value of m for which the line is a tangent to the curve, and find
the coordinates of the point where the tangent touches the curve.
[5]
E
ii
For the case where m = 1, the curve and the line intersect at the points A and B.
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q7 June 2013
12 i
Express 2 x 2 − 4x + 1 in the form a ( x + b )2 + c and hence state the coordinates of the minimum point, A,
on the curve y = 2 x 2 − 4x + 1.
[4]
PL
The line x − y + 4 = 0 intersects the curve y = 2 x 2 − 4x + 1 at the points P and Q.
It is given that the coordinates of P are (3, 7).
ii Find the coordinates of Q.
[3]
iii Find the equation of the line joining Q to the mid-point of AP.
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2011
SA
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Original material © Cambridge University Press 2017
E
PL
SA
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Chapter 2
Functions
In this chapter you will learn how to:
■
■
■
■
■
understand the terms function, domain, range, one-one function, inverse function and
composition of functions
identify the range of a given function in simple cases, and find the composition of two given
functions
determine whether or not a given function is one-one, and find the inverse of a one-one
function in simple cases
illustrate in graphical terms the relation between a one-one function and its inverse
understand and use the transformations of the graph y = f( x ) given by y = f( x ) + a,
y = f( x + a ), y = a f( x ), y = f( ax ) and simple combinations of these.
Original material © Cambridge University Press 2017
33
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
PREREQUISITE KNOWLEDGE
What you should be able to do
Check your skills
IGCSE / O Level Mathematics
Find an output for a given
function.
1 If f( x ) = 3x − 2 , find f(4).
IGCSE / O Level Mathematics
Find a composite function.
2 If f( x ) = 2 x + 1 and
g( x ) = 1 − x, find fg( x ).
IGCSE / O Level Mathematics
Find the inverse of a simple
function.
3 If f( x ) = 5x + 4, find f −1( x ).
Chapter 1
Complete the square.
4 Express 2 x 2 − 12 x + 5 in the
form a ( x + b )2 + c.
PL
E
Where it comes from
Why do we study functions?
At IGCSE level, you learnt how to interpret expressions as functions with inputs and
outputs and find simple composite functions and simple inverse functions.
There are many situations in the real world that can be modelled as functions. Some
examples are:
●●
●●
●●
the temperature of a hot drink as it cools over time
the height of a valve on a bicycle tyre as the bicycle travels along a horizontal road
the depth of water in a conical container as it is filled from a tap
the number of bacteria present after the start of an experiment.
SA
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34
●●
Modelling these situations using appropriate functions enables us to make predictions
about real-life situations such as: How long will it take for the number of bacteria to exceed
5 billion?
In this chapter we will develop a deeper understanding of functions and their special
properties.
WEB LINK
The Thinking
about Functions
and Combining
Functions resources
on the Underground
Mathematics website.
2.1 Definition of a function
A function is a relation that uniquely associates members of one set with members of
another set.
An alternative name for a function is a mapping.
A function can be either a one-one function or a many-one function.
The function x x + 2 where x ∈ , is an example of a one-one function.
f(x)
y = x+2
O
x
Original material © Cambridge University Press 2017
TIP
x ∈ means that x
belongs to the set of
real numbers.
Chapter 2: Functions
A one-one function has one output value for each input value. Equally important is the
fact that for each output value appearing there is only one input value resulting in this
output value.
We can write this function as f : x x + 2 for x ∈  or f( x ) = x + 2 for x ∈ .
f : x x + 2 is read as ‘the function f is such that x is mapped to x + 2’ or ‘f maps x
to x + 2’.
E
f( x ) is the output value of the function f when the input value is x. For example, when
f( x ) = x + 2, f(5) = 5 + 2 = 7.
The function x x 2 where x ∈ , is a many-one function.
f (x)
PL
y = x2
O
x
A many-one function has one output value for each input value but each output value can
have more than one input value.
SA
M
We can write this function as f : x x 2 for x ∈ or f( x ) = x 2 for x ∈ .
f : x x 2 is read as ‘the function f is such that x is mapped to x 2 ’ or ‘f maps x to x 2 ’.
If we now consider the graph of y2 = x :
y
O
y2 = x
x
We can see that the input value shown has two output values. This means that this relation
is not a function.
The set of input values for a function is called the domain of the function.
When defining a function, it is important to also specify its domain.
The set of output values for a function is called the range (or codomain) of the function.
Original material © Cambridge University Press 2017
35
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
WORKED EXAMPLE 2.1
f( x ) = 5 − 2 x for x ∈ , −4 < x < 5
a Write down the domain of the function f .
b Sketch the graph of the function f .
E
c Write down the range of the function f .
Answer
a The domain is −4 < x < 5 .
b The graph of y = 5 − 2 x is a straight line with gradient −2 and y-intercept 5.
When x = −4, y = 5 − 2( − 4) = 13 .
PL
When x = 5, y = 5 − 2(5) = −5.
f(x)
Range
(–4, 13)
O
x
(5, –5)
36
SA
M
Domain
c The range is −5 < f( x ) < 13.
WORKED EXAMPLE 2.2
The function f is defined by f( x ) = ( x − 3)2 + 8 for −1 < x < 9.
Sketch the graph of the function.
Find the range of f .
Answer
f( x ) = ( x − 3)2 + 8 is a positive quadratic function so the graph
will be of the form
.
( x − 3)2 + 8
The minimum value of the expression is 0 + 8 = 8 and this
minimum occurs when x = 3.
The circled part of the expression is a
square so it will always be > 0 .
The smallest value it can be is 0.
This occurs when x = 3.
So the function f( x ) = ( x − 3)2 + 8 will have a minimum
point at the point (3, 8).
When x = −1, y = ( −1 − 3)2 + 8 = 24 .
When x = 9, y = (9 − 3)2 + 8 = 44.
Original material © Cambridge University Press 2017
Chapter 2: Functions
y
Range
(9, 44)
E
(–1, 24)
(3, 8)
O
x
Domain
PL
The range is 8 < f( x ) < 44.
EXERCISE 2A
1 Which of these graphs represent functions? If the graph represents a function,
state whether it is a one-one function or a many-one function.
c
e
y = x2 − 3
for x ∈ y = 2 x 3 − 1 for x ∈ 10
y=
for x ∈ , x > 0
x
d
y = 2x
for x ∈ f
y = 3x 2 + 4 for x ∈ , x > 0
y=
h
y 2 = 4x
for x ∈ , x > 0
x
SA
M
g
x ∈ , x > 0 is
sometimes shortened
to just x > 0.
b
a y = 2 x − 3 for x ∈ TIP
for x ∈ 2 a Represent on a graph the function:
2
 9 − x for x ∈ , −3 < x < 2
x
 2 x + 1 for x ∈ , 2 < x < 4
b State the nature of the function.
3 a Represent on a graph the relation:
2
 x + 1 for 0 < x < 2
y=
 2 x − 3 for 2 < x < 4
b Explain why this relation is not a function.
4 State the domain and range for the functions represented by these two graphs.
a
y
b
(1, 8)
(–2, 20)
y
y = 7 + 2x – x2
y = 2x3 + 3x2 – 12x
(–1, 4)
O
x
(2, 4)
(–3, 9)
O
(5, –8)
x
(1, –7)
Original material © Cambridge University Press 2017
37
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
5 Find the range for each of these functions.
a
f( x ) = x + 4
for x > 8
b f( x ) = 2 x − 7 for −3 < x < 2
c
f( x ) = 7 − 2 x for −1 < x < 4
e
f( x ) = 2 x for −5 < x < 4
d f : x 2 x 2 for 1 < x < 4
12
f f( x ) =
for 1 < x < 8
x
6 Find the range for each of these functions.
f( x ) = x 2 − 2 for x ∈ b
f : x x 2 + 3 for −2 < x < 5
c
f( x ) = 3 − 2 x 2 for x < 2
d f( x ) = 7 − 3x 2 for −1 < x < 2
7 Find the range for each of these functions.
a
f( x ) = ( x − 2)2 + 5
c
f : x 8 − ( x − 5)2 for 4 < x < 10 d
E
a
1
2
for x > 4
b f( x ) = (2 x − 1)2 − 7 for x >
for x > 2
f( x ) = 1 + x − 4
8 Express each function in the form a ( x + b ) + c , where a, b and c are constants
and hence state the range of each function.
a
PL
2
f( x ) = x 2 + 6x − 11 for x ∈ b f( x ) = 3x 2 − 10 x + 2 for x ∈ 9 Express each function in the form a − b( x + c )2 , where a, b and c are constants
and hence state the range of each function.
a
f( x ) = 7 − 8x − x 2 for x ∈ b f( x ) = 2 − 6x − 3x 2 for x ∈ 10 a Represent, on a graph, the function:
2
for
 3 − x
f( x ) = 
 3x − 7 for
0<x<2
2<x<4
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38
b Find the range of the function.
11 The function f : x x 2 + 6x + k , where k is a constant is defined for x ∈ .
Find the range of f in terms of k.
12 The function g : x 5 − ax − 2 x 2 , where a is a constant is defined for x ∈ .
Find the range of g in terms of a.
13 f( x ) = x 2 − 2 x − 3 for x ∈ , − a < x < a
If the range of the function f is −4 < f( x ) < 5, find the value of a.
14 f( x ) = x 2 + x − 4 for x ∈ , a < x < a + 3
If the range of the function f is −2 < f( x ) < 16, find the possible values of a.
15 f( x ) = 2 x 2 − 8x + 5 for x ∈ , 0 < x < k
a Express f( x ) in the form a ( x + b )2 + c .
b State the value of k for which the graph of y = f( x ) has a line of symmetry.
c
For your value of k from part b, find the range of f .
16 Find the largest possible domain for each function and state the corresponding
range.
f( x ) = 3x − 1
1
d f( x ) =
x
a
b f( x ) = x 2 + 2
1
e f( x ) =
x−2
c
f( x ) = 2 x
f
f( x ) =
x−3−2
Original material © Cambridge University Press 2017
Chapter 2: Functions
2.2 Composite functions
Most functions that we meet can be described as combinations of two or more functions.
For example, the function x 3x − 7 is the function ‘multiply by 3 and then subtract 7’.
It is a combination of the two functions g and f where:
g : x 3x
(the function ‘multiply by 3’)
f :xx−7
(the function ‘subtract 7’)
g
f
g(x)
fg(x)
fg
PL
x
E
So, x 3x − 7 can be described as the function ‘first do g, then do f ’.
When one function is followed by another function, the resulting function is called a
composite function.
KEY POINT 2.1
fg( x ) means the function g acts on x first, then f acts on the result.
SA
M
Three important points to remember about composite functions:
KEY POINT 2.2
fg only exists if the range of g is contained within the domain of f .
In general, fg( x ) ≠ gf( x ).
ff( x ) means you apply the function f twice.
EXPLORE 2.1
f( x ) = 2 x − 5 for x ∈ g( x ) = 3x − 1 for x ∈ Three students are asked to find the composite function gf( x ).
Here are their solutions.
Student A
Student B
Student C
gf( x ) = (3x − 1)(2x − 5)
gf( x ) = 2(3x − 1) − 5
= 6x − 7
gf( x ) = 3(2x − 5) − 1
= 6x − 16
= 6x − 17x + 5
2
Discuss these solutions with your classmates.
Which student is correct? What error has each of the other students made?
Original material © Cambridge University Press 2017
39
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
WORKED EXAMPLE 2.3
f( x ) = ( x − 4)2 − 1 for x ∈ g( x ) =
Find fg(4).
2x + 3
for x ∈ , x > 2
x−2
Answer
11
fg(4) = f  
 2 
2
11
=
− 4 − 1
 2

f is the function ‘subtract 4, square and then
subtract 1’.
WORKED EXAMPLE 2.4
f( x ) = 2 x + 3 for x ∈ a
fg( x ) = f( x 2 − 1)
40
g( x ) = x 2 − 1 for x ∈ fg( x ) b
Answer
a
PL
= 1 14
Find:
2(4) + 3 11
= .
4−2
2
E
g acts on 4 first and g(4) =
2
= 2( x − 1) + 3
gf( x ) c
ff( x )
g acts on x first and g( x ) = x 2 − 1.
f is the function ‘double and add 3’.
SA
M
= 2x2 + 1
b
gf( x ) = g(2 x + 3)
f acts on x first and f( x ) = 2 x + 3 .
2
= (2 x + 3) − 1
g is the function ‘square and subtract 1’.
= 4x 2 + 12 x + 9 − 1
= 4x 2 + 12 x + 8
ff( x ) = f(2 x + 3)
c
f is the function ‘double and add 3’.
= 2(2 x + 3) + 3
= 4x + 9
WORKED EXAMPLE 2.5
f :x
Find
a
5
for x ∈ , x ≠ 2
x−2
fg( x ) g( x ) = 3 − x 2 for x ∈ b
ff( x )
Answer
a
fg( x ) = f(3 − x 2 )
5
=
(3 − x 2 ) − 2
5
=
1 − x2
g acts on x first and g( x ) = 3 − x 2 .
f is the function ‘subtract 2 and then divide into 5’.
Original material © Cambridge University Press 2017
Chapter 2: Functions
5 
ff ( x ) = f 
 x−2
5
=
5
−2
x−2
=
5( x − 2)
5 − 2( x − 2)
=
5x − 10
9 − 2x
Multiply numerator and denominator by ( x − 2).
E
b
f( x ) = x 2 + 4x for x ∈ PL
WORKED EXAMPLE 2.6
g( x ) = 3x − 1 for x ∈ Find the values of k for which the equation fg( x ) = k has real solutions.
Answer
fg( x ) = (3x − 1)2 + 4(3x − 1)
2
= 9x + 6x − 3
When fg( x ) = k,
9x 2 + 6x − 3 = k
9x 2 + 6x + ( −3 − k ) = 0
Expand brackets and simplify.
Rearrange and simplify.
b2 − 4ac > 0
SA
M
For real solutions:
62 − 4 × 9 × ( −3 − k ) > 0
144 + 36 k > 0
k > −4
EXERCISE 2B
1 f( x ) = x 2 + 6 for x ∈ g( x ) =
Find:
a
b
fg(6)
x + 3 − 2 for x ∈ , x > −3
gf(4)
2 h : x x + 5 for x ∈ , x . 0 k : x c
ff( −3)
c
x x + 10
x for x ∈ , x . 0
Express each of the following in terms of h and/or k.
a
x
x +5
b
x
x+5
3 f( x ) = ax + b for x ∈ Given that f(5) = 3 and f(3) = −3:
a find the value of a and the value of b
b solve the equation ff( x ) = 4.
4 f : x 2 x + 3 for x ∈ g : x 12
for x ∈ , x ≠ 1
1− x
a Find gf( x ). b
Solve the equation gf( x ) = 2 .
Original material © Cambridge University Press 2017
41
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
5 g( x ) = x 2 − 2 for x ∈ h( x ) = 2 x + 5 for x ∈ a Find gh( x ).
b Solve the equation gh( x ) = 14 .
6
f( x ) = x 2 + 1 for x ∈ g( x ) =
Solve the equation fg( x ) = 5 .
3
for x ∈ , x ≠ 2
x−2
2
for x ∈ , x ≠ −1 h( x ) = ( x + 2)2 − 5 for x ∈ x +1
Solve the equation hg( x ) = 11.
E
7 g( x ) =
x +1
2x + 3
for x ∈ g : x for x ∈ , x ≠ 1
2
x −1
Solve the equation gf( x ) = 1.
8 f:x
PL
x +1
for x ∈ , x . 0
2x + 5
Find an expression for ff( x ) , giving your answer as a single fraction in its simplest form.
9 f( x ) =
10 f : x x 2 for x ∈ g : x x + 1 for x ∈ Express each of the following as a composite function, using only f and/or g.
x ( x + 1)2
d
x x4
b
x x2 + 1
c
xx+2
e
x x2 + 2x + 2
f
x x 4 + 2x2 + 1
11 f( x ) = x 2 − 3x for x ∈ g( x ) = 2 x + 5 for x ∈ Show that the equation gf( x ) = 0 has no real solutions.
SA
M
42
a
2
for x ∈ , x ≠ 0
x
Find the values of k for which the equation fg( x ) = x has two equal roots.
12 f( x ) = k − 2 x for x ∈ g( x ) =
13 f( x ) = x 2 − 3x for x ∈ g( x ) = 2 x − 5 for x ∈ Find the values of k for which the equation gf( x ) = k has real solutions.
x+5
1
for x ∈ , x ≠
2x − 1
2
Show that ff( x ) = x.
14 f( x ) =
15 f( x ) = 2 x 2 + 4x − 8 for x ∈ , x > k
a Express 2 x 2 + 4x − 8 in the form a ( x + b )2 + c.
b Find the least value of k for which the function is one-one.
16 f( x ) = x 2 − 2 x + 4 for x ∈ a Find the set of values of x for which f( x ) > 7 .
b Express x 2 − 2 x + 4 in the form ( x − a )2 + b.
c
Write down the range of f .
17 f( x ) = x 2 − 5x for x ∈ g( x ) = 2 x + 3 for x ∈ a Find fg( x ). b Find the range of the function fg( x ).
Original material © Cambridge University Press 2017
Chapter 2: Functions
2
for x ∈ , x ≠ −1
x +1
a Find ff( x ) and state the domain of this function.
18 f( x ) =
b Show that if f( x ) = ff( x ) then x 2 + x − 2 = 0.
PS
Find the values of x for which f( x ) = ff( x ).
19
P( x ) = x 2 − 1 for x ∈ R( x ) =
1
for x ∈ , x ≠ 0
x
Q( x ) = x + 2 for x ∈ S( x ) =
E
c
x + 1 − 1 for x ∈ , x > −1
Functions P, Q, R and S are composed in some way to make a new function, f( x ).
2.3 Inverse functions
PL
For each of the following, write f( x ) in terms of the functions P, Q, R and/or S, and state the domain and
range for each composite function.
1
b f( x ) = x 2 + 1
c f( x ) = x
d f( x ) = 2 + 1
a f( x ) = x 2 + 4x + 3
x
1
f f( x ) = x − 2 x + 1 + 1
g f( x ) = x − 1
e f( x ) =
x+4
The inverse of a function f( x ) is the function that undoes what f( x ) has done.
f(x)
We write the inverse of the function f( x ) as f −1( x ) .
SA
M
KEY POINT 2.3
x
y
ff −1( x ) = f −1f( x ) = x
The domain of f −1( x ) is the range of f( x ).
The range of f −1( x ) is the domain of f( x ).
It is important to remember that not every function has an inverse.
KEY POINT 2.4
An inverse function f −1( x ) exists if, and only if, the function f( x ) is a one-one mapping.
You should already know how to find the inverse function of some simple one-one
mappings.
We want to find the function f −1( x ), so if we write y = f −1( x ), then f( y ) = f(f −1( x )) = x,
because f and f −1 are inverse functions. So if we write x = f( y ) and then rearrange it to get
y = … , then the right hand side will be f −1( x ).
Original material © Cambridge University Press 2017
43
f –1(x)
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
We find the inverse of the function f( x ) = 3x − 1 by following the steps:
Step 1: Write the function as y =
Step 2: Interchange the x and y variables.
Step 3: Rearrange to make y the subject.
y
yy
x
x
x
yy
y
=
=
=
=
=
=
=
=
=
33x
1
x−
−
3x
− 11
33 yy −
1
3y −
− 11
x
+
x
+ 111
x+
33
3
x +1
.
3
If f and f −1 are the same function, then f is called a self-inverse function.
1
1
for x ≠ 0 , then f −1( x ) =
for x ≠ 0 .
For example, if f( x ) =
x
x
1
for x ≠ 0 is a self-inverse function.
So f( x ) =
x
PL
E
Hence, if f( x ) = 3x − 1 then f −1( x ) =
EXPLORE 2.2
The diagram shows the function f( x ) = ( x − 2)2 + 1 for x ∈ .
Discuss the following questions with your classmates:
f(x)
f(x) = (x – 2)2 + 1
1 What type of mapping is this function?
2 What are the coordinates of the vertex of the parabola?
3 What is the domain of the function?
O
4 What is the range of the function?
SA
M
44
x
5 Does this function have an inverse?
6 If f has an inverse, what is it? If not, then how could you change the domain of f so that the function does have
an inverse?
WORKED EXAMPLE 2.7
f( x ) =
x + 2 − 7 for x ∈ , x > −2
a Find an expression for f −1( x ) .
b Solve the equation f −1( x ) = f(62) .
Answer
a
f( x ) =
x+2 −7
Step 1: Write the function as y =
y=
x+2 −7
Step 2: Interchange the x and y variables.
x=
y+2 −7
Step 3: Rearrange to make y the subject.
x+7 =
y+2
( x + 7)2 = y + 2
y = ( x + 7)2 − 2
Original material © Cambridge University Press 2017
Chapter 2: Functions
b
f −1( x ) = ( x + 7)2 − 2
f(62) =
62 + 2 − 7 = 1
( x + 7)2 − 2 = 1
( x + 7)2 = 3
x+7 = ± 3
x = −7 − 3 or x = −7 + 3
E
x = −7 ± 3
The range of f is f( x ) > −7 so the domain of f −1 is x > −7.
WORKED EXAMPLE 2.8
PL
Hence the only solution of f −1( x ) = f(62) is x = −7 + 3 .
f( x ) = 5 − ( x − 2)2 for x ∈ , k < x < 6
a State the smallest value of k for which f has an inverse.
b For this value of k find an expression for f −1( x ) , and state the domain and range of f −1 .
Answer
y
(2, 5)
45
a The vertex of the graph of y = 5 − ( x − 2)2 is at the point (2, 5).
y=
SA
M
When x = 6, y = 5 − 42 = −11
For the function f to have an inverse it must be a one-one function.
O
5 – (x – 2)2
x
Hence the smallest value of k is 2.
b
f( x ) = 5 − ( x − 2)2
(6, –11)
Step 1: Write the function as y =
y = 5 − ( x − 2)2
Step 2: Interchange the x and y variables.
x = 5 − ( y − 2)2
Step 3: Rearrange to make y the subject.
( y − 2)2 = 5 − x
y−2 = 5−x
y =2+ 5−x
Hence, f −1( x ) = 2 + 5 − x
The domain of f −1 is the same as the range of f .
Hence, the domain of f −1 is −11 < x < 5 .
The range of f −1 is the same as the domain of f .
Hence, the range of f −1 is 2 < f −1 ( x ) < 6.
Original material © Cambridge University Press 2017
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXERCISE 2C
1 Find an expression for f −1( x ) for each of the following functions.
f( x ) = 5x − 8 for x ∈ b
c
f( x ) = ( x − 5)2 + 3 for x ∈ , x > 5
d
e
f( x ) =
x+7
for x ∈ , x ≠ −2
x+2
f : x x 2 + 4x for x ∈ , x > −2
a State the domain and range of f −1.
b Find an expression for f −1( x ) .
5
for x ∈ , x > 2
2x + 1
a Find an expression for f −1( x ).
f:x
f( x ) = ( x − 2)3 − 1 for x ∈ , x > 2
PL
3
f
f( x ) = x 2 + 3 for x ∈ , x > 0
8
f( x ) =
for x ∈ , x ≠ 3
x−3
E
2
a
b Find the domain of f −1.
4
f : x ( x + 1)3 − 4 for x ∈ , x > 0
a Find an expression for f −1( x ) .
b Find the domain of f −1.
5
g : x 2 x 2 − 8x + 10 for x ∈ , x > 3
a Explain why g has an inverse.
SA
M
46
b Find an expression for g −1( x ).
6
f : x 2 x 2 + 12 x − 14 for x ∈ , x > k
a Find the least value of k for which f is one-one.
b Find an expression for f −1( x ).
7
f : x x 2 − 6x for x ∈ a Find the range of f.
b State, with a reason, whether f has an inverse.
8
f( x ) = 9 − ( x − 3)2 for x ∈ , k < x < 7
a State the smallest value of k for which f has an inverse.
b For this value of k:
i
find an expression for f −1( x )
ii
state the domain and range of f −1.
Original material © Cambridge University Press 2017
Chapter 2: Functions
9
y
y=
5x – 1
x
x
O
5x − 1
for x ∈ , 0 , x < 3.
x
E
The diagram shows the graph of y = f −1( x ) , where f −1( x ) =
a Find an expression for f( x ).
b State the domain of f.
10 f( x ) = 3x + a for x ∈ g( x ) = b − 5x for x ∈ Given that gf( −1) = 2 and g −1(7) = 1, find the value of a and the value of b.
3
for x ∈ , x ≠ 2
2x − 4
PL
11 f( x ) = 3x − 1 for x ∈ g( x ) =
a Find expressions for f −1( x ) and g −1( x ).
b Show that the equation f −1( x ) = g −1( x ) has two real roots.
12 f : x (2 x − 1)3 − 3 for x ∈ , 1 < x < 3
a Find an expression for f −1( x ) .
b Find the domain of f −1.
13 f : x x 2 − 10 x for x ∈ , x > 5
SA
M
a Express f( x ) in the form ( x − a )2 − b.
b Find an expression for f −1( x ) and state the domain of f −1.
1
for x ∈ , x ≠ 1
x −1
a Find an expression for f −1( x ).
14 f( x ) =
b Show that if f( x ) = f −1( x ) , then x 2 − x − 1 = 0 .
c
Find the values of x for which f( x ) = f −1( x ) .
Give your answer in surd form.
15 Determine which of the following functions are self-inverse functions.
1
2x + 1
a f( x ) =
for x ∈ , x ≠ 3
b f( x ) =
for x ∈ , x ≠ 2
x−2
3−x
3x + 5
3
for x ∈ , x ≠
c f( x ) =
4x − 3
4
16 f : x 3x − 5 for x ∈ g : x 4 − 2 x for x ∈ a Find an expression for (fg) −1( x ).
b Find expressions for:
i
c
f −1 g −1( x )
ii
g −1 f −1( x ) .
Comment on your results in part b.
Investigate if this is true for other functions.
Original material © Cambridge University Press 2017
47
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
2.4 The graph of a function and its inverse
Consider the function defined by f( x ) = 2 x + 1 for x ∈ , −4 < x < 2 .
f( −4) = −7 and f(2) = 5.
y
The domain of f is −4 < x < 2 and the range is −7 < f( x ) < 5.
x −1
.
The inverse of this function is f −1( x ) =
2
The domain of f −1 is the same as the range of f.
(2, 5)
y=x
E
(5, 2)
Hence, the domain of f −1 is −7 < x < 5.
x
f–1
The range of f −1 is the same as the domain of f.
(–7, –4)
Hence, the range of f −1 is −4 < f −1( x ) < 2.
The representation of f and f −1 on the same graph can be seen in the diagram opposite.
f
PL
(–4, –7)
It is important to note that the graphs of f and f −1 are reflections of each other in the line
y = x. This is true for each one-one function and its inverse functions.
KEY POINT 2.5
The graphs of f and f −1 are reflections of each other in the line y = x.
This is because: ff −1( x ) = x = f −1 f( x )
When a function f is self-inverse, the graph of f will be symmetrical about the line y = x.
WORKED EXAMPLE 2.9
SA
M
48
f( x ) = ( x − 1)2 − 2 for x ∈ , 1 < x < 4
On the same axes, draw the graph of f and the graph of f −1.
Answer
y = ( x − 1)2 − 2
When x = 4, y = 7.
The function is one-one, so the inverse function exists.
The circled part of the expression is a square so it
will always be > 0. The smallest value it can be is 0.
This occurs when x = 1. The vortex is at the
point (1, −2) .
y
8
y
8
y=x
f
f
6
6
Reflect f in y = x
4
2
–2
O
–2
2
4
6
8 x
f –1
4
2
–2
O
–2
Original material © Cambridge University Press 2017
2
4
6
8 x
Chapter 2: Functions
WORKED EXAMPLE 2.10
2x + 7
for x ∈ , x ≠ 2
x−2
a Find an expression for f −1( x ) .
f:x
b State what your answer to part a tells you about the symmetry of the graph of y = f( x ).
a
f :x
E
Answer
2x + 7
x−2
Step 1: Write the function as y =
2x + 7
x−2
x=
2y + 7
y−2
PL
Step 2: Interchange the x and y variables.
y=
xy − 2 x = 2 y + 7
Step 3: Rearrange to make y the subject.
y( x − 2) = 2 x + 7
2x + 7
y=
x−2
Hence f −1( x ) =
f −1( x ) = f( x ) , so the function f is self-inverse.
SA
M
b
2x + 7
.
x−2
The graph of y = f( x ) is symmetrical about the line y = x.
EXPLORE 2.3
y
y
y=x
f
g–1
y=x
g
f–1
O
x
O
x
Ali states that:
The diagrams show the functions f and g together with their inverse functions f −1 and g−1.
Is Ali correct?
Explain your answer.
Original material © Cambridge University Press 2017
49
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXERCISE 2D
1 On a copy of each grid, draw the graph of f −1( x ) if it exists.
y
6
a
y
6
b
f
4
f
–4
–2
O
2
2
6 x
4
PL
y
4
2
1
1
SA
M
O
f
3
f
2
2
6 x
4
5
4
50
2
6
5
3
O
–2
–6
d
y
6
–4
–4
–4
c
–6
–2
–2
–6
E
2
–6
4
1
2
3
4
5
6
O
x
1
2
3
4
5
6
x
f : x 2 x − 1 for x ∈ , −1 < x < 3
a Find an expression for f −1( x ).
b State the domain and range of f −1.
c
Sketch, in a single diagram, the graphs of y = f( x ) and y = f −1( x ) , making clear the relationship between
the graphs.
3 The diagram shows the graph of y = f( x ) , where f( x ) =
a State the range of f .
4
for x ∈ , x > 0.
x+2
y
b Find an expression for f −1( x ) .
c
State the domain and range of f −1.
d On a copy of the diagram, sketch the graph of y = f −1( x ), making clear the
relationship between the graphs.
f
O
4 For each of the following functions, find an expression for f −1( x ) and hence decide if the graph of y = f( x ) is
symmetrical about the line y = x.
x+5
2x − 3
1
for x ∈ , x ≠
b f( x ) =
for x ∈ , x ≠ 5
a f( x ) =
2x − 1
x−5
2
c
f( x ) =
3
3x − 1
for x ∈ , x ≠
2x − 3
2
d
f( x ) =
4
4x + 5
for x ∈ , x ≠
3x − 4
3
Original material © Cambridge University Press 2017
x
Chapter 2: Functions
P
x+a
1
for x ∈ , x ≠ , where a and b are constants.
bx − 1
b
Prove that this function is self-inverse.
ax + b
d
for x ∈ , x ≠ − , where a, b, c and d are constants.
b g( x ) =
cx + d
c
Find the condition for this function to be self-inverse.
5 a
f( x ) =
E
2.5 Transformations of functions
EXPLORE 2.4
PL
At IGCSE level you met various transformations that can be applied to two-dimensional
shapes. These included translations, reflections, rotations and enlargements. In this section
we will learn how translations, reflections and stretches (and combinations of these) can be
used to transform the graph of a function.
1 a Use graphing software to draw the graphs of y = x 2, y = x 2 + 2 and y = x 2 − 3.
Discuss your observations with your classmates and explain how the second and third graphs could be
obtained from the first graph.
b Repeat part a using the graphs y =
c Repeat part a using the graphs y =
x, y =
x + 1 and y =
x − 2.
12
12
12
,y=
+ 5 and y =
− 4.
x
x
x
d Can you generalise your results?
2 a
Use graphing software to draw the graphs of y = x 2, y = ( x + 2)2 and y = ( x − 5)2 .
SA
M
Discuss your observations with your classmates and explain how the second and third graphs could be
obtained from the first graph.
b Repeat part a using the graphs y = x 3, y = ( x + 1)3 and y = ( x − 4)3 .
c Can you generalise your results?
3 a Use graphing software to draw the graphs of y = x 2 and y = − x 2 .
Discuss your observations with your classmates and explain how the second graph could be obtained from
the first graph.
b Repeat part a using the graphs y = x 3 and y = − x 3.
c Repeat part a using the graphs y = 2 x and y = −2 x .
d Can you generalise your results?
4 a
Use graphing software to draw the graphs of y = 5 + x and y = 5 − x.
Discuss your observations with your classmates and explain how the second graph could be obtained from
the first graph.
b Repeat part a using the graphs y =
2 + x and y =
2 − x.
c Can you generalise your results?
5 aUse graphing software to draw the graphs of y = x 2 and y = 2 x 2 and y = (2 x )2.
Discuss your observations with your classmates and explain how the second graph could be obtained from
the first graph.
b Repeat part a using the graphs y =
x , y = 2 x and y =
2x .
c Repeat part a using the graphs y = 3 , y = 2 × 3 and y = 32 x .
x
x
d Can you generalise your results?
Original material © Cambridge University Press 2017
51
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Translations
y
12
The diagram shows the graphs of two functions that differ only by a constant.
y = x2 – 2x + 4
2
y = x − 2x + 1
10
y = x2 − 2x + 4
8
When the x-coordinates on the two graphs are the same ( x = x ) the
y-coordinates differ by 3 ( y = y + 3).
y = x2 – 2x + 1
E
6
This means that the two curves have exactly the same shape but that they are
separated by 3 units in the positive y direction.
4
Hence the graph of y = x 2 − 2 x + 4 is a translation of the graph of y = x 2 − 2 x + 1
2
–2
KEY POINT 2.6
 0
.
 a 
The graph of y = f( x ) + a is a translation of the graph y = f( x ) by the vector 
Now consider the two functions:
y = x2 − 2x + 1
y = ( x − 3)2 − 2( x − 3) + 1
SA
M
52
We obtain the second function by replacing x by x − 3 in the first function.
The graphs of these two functions are:
y
8
y = x2 – 2x + 1
y = (x – 3)2 – 2 (x – 3) + 1
6
4
3
2
–2
O
2
4
O
PL
 0
by the vector   .
 3
3
6
x
The curves have exactly the same shape but this time they are separated by 3 units in the
positive x direction.
You may be surprised that the curve has moved in the positive x direction. Note, however,
that a way of obtaining y = y is to have x = x − 3 or equivalently x = x + 3. This means
that the two curves are at the same height when the red curve is 3 units to the right of the
blue curve.
Original material © Cambridge University Press 2017
2
4
x
Chapter 2: Functions
Hence the graph of y = ( x − 3)2 − 2( x − 3) + 1 is a translation of the graph of
 3
y = x 2 − 2 x + 1 by the vector   .
 0
KEY POINT 2.7
 a
.
 0 
E
The graph of y = f( x − a ) is a translation of the graph y = f( x ) by the vector 
Combining these two results gives:
PL
KEY POINT 2.8
 
The graph of y = f( x − a ) + b is a translation of the graph y = f( x ) by the vector a .
 b 
WORKED EXAMPLE 2.11
The graph of y = x 2 + 5x is translated 2 units to the right. Find the equation of the resulting graph. Give your
answer in the form y = ax 2 + bx + c.
Answer
y = x 2 + 5x
SA
M
Replace all occurrences of x by x − 2.
2
y = ( x − 2) + 5( x − 2)
Expand and simplify.
y = x2 + x − 6
WORKED EXAMPLE 2.12
The graph of y =
Answer
y=
2x
y=
2( x + 5) + 3
y=
2 x + 10 + 3
 −5 
2 x is translated by the vector   . Find the equation of the resulting graph.
 3
Replace x by x + 5 , and add 3 to the resulting function.
Original material © Cambridge University Press 2017
53
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXERCISE 2E
 0
after translation by  
 4
b
y=5 x
 0
after translation by  
 −2 
c
y = 7x2 − 2x
 0
after translation by  
 1
d
y = x2 − 1
 0
after translation by  
 2
e
y=
2
x
 −5 
after translation by  
 0
f
y=
x
x +1
 3
after translation by  
 0
g
y = x2 + x
 −1 
after translation by  
 0
h
y = 3x 2 − 2
 2
after translation by  
 3
PL
y = 2x2
2 Find the translation that transforms the graph.
a
y = x 2 + 5x − 2 to the graph y = x 2 + 5x + 2 ,
SA
M
54
a
E
1 Find the equation of each graph after the given transformation.
b
y = x3 + 2x2 + 1
to the graph y = x 3 + 2 x 2 − 4 ,
c
y = x 2 − 3x 6
y = x + x
y = 2 x + 5 5
y = 2 − 3x x
to the graph y = ( x + 1)2 − 3( x + 1),
6
,
to the graph y = x − 2 +
x−2
to the graph y = 2 x + 3 ,
5
to the graph y =
− 3x + 10 .
( x − 2)2
d
e
f
3 The diagram shows the graph of y = f( x ).
Sketch the graphs of each of the following functions.
a
y = f( x ) − 4
b
y = f ( x − 2)
c
y = f ( x + 1) − 5
y
4
3
2
1
–4 –3 –2 –1 O
–1
–2
–3
–4
y = f(x)
1 2 3 4 x
4 a
On the same diagram sketch the graphs of y = 2 x and y = 2 x + 2.
b
 0
y = 2 x can be transformed to y = 2 x + 2 by a translation of   .
 a
Find the value of a.
c
 b
y = 2 x can be transformed to y = 2 x + 2 by a translation of   .
 0
Find the value of b.
Original material © Cambridge University Press 2017
Chapter 2: Functions
5 A cubic graph has equation y = ( x + 3 ) ( x − 2 ) ( x − 5 ) .
 2
Write in a similar form the equation of the graph after a translation of   .
 0
 1
6 The graph of y = x 2 − 4x + 1 is translated by the vector   .
 2
WEB LINK
Try the Between the
lines resource on
the Underground
Mathematics website.
Find, in the form y = ax 2 + bx + c, the equation of the resulting graph.
 2
7 The graph of y = ax 2 + bx + c is translated by the vector   .
 −5 
2
The resulting graph is y = 2 x − 11x + 10 . Find the value of a, the value of b and
the value of c.
E
PS
y
6
PL
2.6 Reflections
The diagram shows the graphs of the two functions:
4
y = x2 − 2x + 1
y = −( x 2 − 2 x + 1)
2
When the x-coordinates on the two graphs are the same ( x = x ) the y-coordinates
are negative of each other ( y = − y ).
–2
O
–2
Hence the graph of y = −( x 2 − 2 x + 1) is a reflection of the graph of y = x 2 − 2 x + 1
in the x-axis.
–6
SA
M
This means that, when the x-coordinates are the same, the red curve is the same
vertical distance from the x-axis as the blue curve but it is on the opposite side of the
x-axis.
KEY POINT 2.9
The graph of y = −f( x ) is a reflection of the graph y = f( x ) in the x-axis.
Now consider the two functions:
y = x2 − 2x + 1
y = ( − x )2 − 2( − x ) + 1
We obtain the second function by replacing x by −x in the first function.
The graphs of these two functions are demonstrated in the diagram.
y
6
5
4
y = (–x)2 – 2(–x) + 1
y = x2 – 2x + 1
3
2
1
–4
–2
y = x2 – 2x + 1
O
2
4
x
Original material © Cambridge University Press 2017
–4
2
4 x
y = –(x2 – 2x + 1)
55
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
The curves are at the same height ( y = y ) when x = − x or equivalently x = − x.
This means that the heights of the two graphs are the same, when the red graph has the
same horizontal displacement from the y-axis as the blue graph but is on the opposite side
of the y-axis.
KEY POINT 2.10
WORKED EXAMPLE 2.13
PL
The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis.
E
Hence the graph of y = ( − x )2 − 2( − x ) + 1 is a reflection of the graph of y = x 2 − 2 x + 1 in
the y-axis.
The quadratic graph y = f( x ) has a minimum at the point (5, −7). Find the coordinates of the vertex and state
whether it is a maximum or minimum of the graph for each of the following graphs.
a
y = − f( x )
Answer
a
y = f( − x )
b
y = − f( x ) is a reflection of y = f( x ) in the x-axis.
The turning point is (5, 7). It is a maximum point.
56
y = f( − x ) is a reflection of y = f( x ) in the y-axis.
SA
M
b
The turning point is ( −5, −7) . It is a minimum point.
EXERCISE 2F
1 The diagram shows the graph of y = g( x ) .
Sketch the graphs of each of the following functions.
a
y = −g( x )
b
y = g( − x )
y
4
3
2
y = g(x)
1
–4 –3 –2 –1 O
–1
1
2
3
–2
–3
–4
2 Find the equation of each graph after the given transformation.
a
y = 5x 2 after reflection in the x-axis.
b
y = 2 x 4 after reflection in the y-axis.
c
y = 2 x 2 − 3x + 1 after reflection in the y-axis.
d
y = 5 + 2 x − 3x 2 after reflection in the x-axis.
Original material © Cambridge University Press 2017
4 x
Chapter 2: Functions
3 Describe the transformation that maps the graph:
y = x 2 + 7 x − 3 onto the graph y = − x 2 − 7 x + 3,
b
y = x 2 − 3x + 4 onto the graph y = x 2 + 3x + 4,
c
y = 2 x − 5x 2 onto the graph y = 5x 2 − 2 x,
d
y = x 3 + 2 x 2 − 3x + 1 onto the graph y = − x 3 − 2 x 2 + 3x − 1.
2.7 Stretches
E
a
y
The diagram shows the graphs of the two functions:
10
×2
y = x2 − 2x − 3
PL
y = 2( x 2 − 2 x − 3)
When the x-coordinates on the two graphs are the same ( x = x ) the
y-coordinate on the red graph is double the y-coordinate on the blue
graph ( y = 2 y ).
y = x 2 – 2x – 3
5
This means that, when the x-coordinates are the same, the red curve is twice the
distance of the blue graph from the x-axis.
–4
Hence the graph of y = 2( x 2 − 2 x − 3) is a stretch of the graph of
2
y = x − 2 x − 3 from the x-axis. We say that it has been stretched with stretch
factor 2 parallel to the y-axis.
SA
M
Note: There are alternative ways of expressing this transformation:
●●
●●
●●
●●
a stretch with scale factor 2 with the line y = 0 invariant
a stretch with stretch factor 2 with the x-axis invariant
a stretch with stretch factor 2 relative to the x-axis
a vertical stretch with stretch factor 2.
KEY POINT 2.11
The graph of y = a f( x ) is a stretch of the graph y = f( x ) with stretch factor a parallel to the
y-axis.
Note: If a , 0 then y = a f( x ) can be considered to be a stretch of
y = f( x ) with a negative scale factor or as a stretch with positive scale
factor followed by a reflection in the x-axis.
Original material © Cambridge University Press 2017
–2
O
–5
2
4
6 x
×2
y = 2(x 2 – 2x – 3)
57
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
Now consider the two functions:
y = x2 − 2x − 3
×1
2
2
y = (2 x ) − 2(2 x ) − 3
y = (2x)2 – 2(2x) – 3
10
We obtain the second function by replacing x by 2x in the first function.
The two curves are at the same height ( y = y ) when x = 2 x or equivalently
1
x = x.
2
This means that the heights of the two graphs are the same when the red graph
has half the horizontal displacement from the y-axis as the blue graph.
5
y = x2 – 2x – 3
E
Hence the graph of y = (2 x )2 − 2(2 x ) − 3 is a stretch of the graph of
×1
2
–4
–2
O
2
4
6
PL
y = x 2 − 2 x − 3 from the y-axis. We say that it has been stretched with stretch
1
factor parallel to the x-axis.
2
–5
KEY POINT 2.12
The graph of y = f( ax ) is a stretch of the graph y = f( x ) with stretch factor
x-axis.
SA
M
58
1
parallel to the
a
WORKED EXAMPLE 2.14
The graph of y = 5 −
1 2
x is stretched with stretch factor 4 parallel to the y-axis.
2
Find the equation of the resulting graph.
Answer
1 2
x
2
4f( x ) = 20 − 2 x 2
Let f( x ) = 5 −
A stretch parallel to the y-axis, factor 4, gives
the function 4f( x ).
The equation of the resulting graph is y = 20 − 2 x 2.
WORKED EXAMPLE 2.15
Describe the single transformation that maps the graph of y = x 2 − 3x − 5 to the graph of y = 4x 2 − 6x − 5.
Answer
Let f( x ) = x 2 − 3x − 5
Express 4x 2 − 6x − 5 in terms of f( x ).
4x 2 − 6x − 5 = (2 x )2 − 3(2 x ) − 5
= f(2 x )
1
The transformation is a stretch parallel to the x-axis with stretch factor .
2
Original material © Cambridge University Press 2017
x
Chapter 2: Functions
EXERCISE 2G
y
6
1 The diagram shows the graph of y = f( x ).
Sketch the graphs of each of the following functions.
a
y = 3f( x )
b
y = f(2 x )
4
2
–4
–2 O
4
6 x
y = f(x)
2
E
–6
–2
–4
–6
PL
2 Find the equation of each graph after the given transformation.
a
y = 3x 2 after a stretch parallel to the y-axis with stretch factor 2
b
y = x 3 − 1 after a stretch parallel to the y-axis with stretch factor 3
1
y = 2 x + 4 after a stretch parallel to the y-axis with stretch factor
2
y = 2 x 2 − 8x + 10 after a stretch parallel to the x-axis with stretch factor 2
1
y = 6x 3 − 36x after a stretch parallel to the x-axis with stretch factor
3
c
d
e
3 Describe the single transformation that maps the graph:
y = x 2 + 2 x − 5 onto the graph y = 4x 2 + 4x − 5
b
y = x 2 − 3x + 2 onto the graph y = 3x 2 − 9x + 6
c
y = 2 x + 1 onto the graph y = 2 x +1 + 2
d
y=
SA
M
a
x − 6 onto the graph y =
3x − 6
2.8 Combined transformations
In this section we will learn how to apply simple combinations of transformations.
The transformations of the graph of y = f( x ) that we have studied so far can each be
categorised as either vertical or horizontal transformations.
Vertical transformations
y = f( x ) + a
 0
translation  
 a
Horizontal transformations
y = f( x + a )
y = −f( x )
reflection in the x-axis
y = a f( x )
vertical stretch, factor a
 −a 
translation 

 0
y = f( − x )
reflection in the y-axis
y = f( ax )
horizontal stretch, factor
When combining transformations care must be taken with the order in which the
transformations are applied.
Original material © Cambridge University Press 2017
1
a
59
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
EXPLORE 2.5
Apply the transformations in the given order to triangle T
and for each question comment on whether the final images
are the same or different.
y
2
1
T
1 Combining two vertical transformations
 0
1
then translate   .
2
 3
b Investigate for other pairs of vertical transformations.
2
3 x
PL
ii Stretch vertically with factor
1
E
 0
1
a i Translate   then stretch vertically with factor .
2
 3
O
2 Combining one vertical and one horizontal transformation
 −2 
a i Reflect in the x-axis then translate   .
 0
 −2 
ii Translate   then reflect in the x-axis.
 0
b I nvestigate for other pairs of transformations where one is vertical and the
other is horizontal.
3 Combining two horizontal transformations
 2
a i Stretch horizontally with factor 2 then translate   .
 0
 2
ii Translate   then stretch horizontally with factor 2.
 0
SA
M
60
b Investigate for other pairs of horizontal transformations.
From the Explore activity, you should have found that:
KEY POINT 2.13
• w
hen two vertical transformations or two horizontal transformations are combined, the order
in which they are applied may affect the outcome.
• w
hen one horizontal and one vertical transformation are combined, the order in which they
are applied does not affect the outcome.
Combining two vertical transformations
We will now consider how the graph of y = f( x ) is transformed to the graph y = a f( x ) + k.
This can be shown in a flow diagram as:
f( x ) →
stretch vertically, factor a
multiply function by a
→ a f( x ) →
 0
translate  
 k
→ a f( x ) + k
add k to the function
Original material © Cambridge University Press 2017
Chapter 2: Functions
This leads to the important result:
KEY POINT 2.14
Vertical transformations follow the ‘normal’ order of operations as used in arithmetic.
Combining two horizontal transformations
f( x ) →
 −c 
translate  
 0
replace x with x + c
→ f( x + c ) →
stretch horizontally, factor
1
b → f( bx + c )
replace x with bx
PL
This leads to the important result:
KEY POINT 2.15
E
Now consider how the graph of y = f( x ) is transformed to the graph y = f( bx + c )
Horizontal transformations follow the opposite order to the ‘normal’ order of operations as used
in arithmetic.
WORKED EXAMPLE 2.16
SA
M
The diagram shows the graph of y = f( x ) .
61
y
6
Sketch the graph of y = 2f( x ) − 3.
4
y = f(x)
2
–6
–4
–2
O
2
6 x
4
–2
–4
–6
Answer
y = 2f( x ) − 3 is a combination of two vertical transformations of y = f( x ) ,
hence the transformations follow the ‘normal’ order of operations.
y
6
Step 1: Sketch the graph y = 2f( x ):
4
Stretch y = f( x ) vertically with stretch factor 2 .
y = 2f(x)
2
–6
–4
–2
O
–2
–4
–6
Original material © Cambridge University Press 2017
2
4
6 x
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Step 2: Sketch the graph y = 2f ( x ) − 3:
y
6
 
Translate y = 2f ( x ) by the vector  0  .
 −3 
4
y = 2f(x) – 3
2
–6
–4
–2
O
2
4
E
–2
–4
–6
PL
WORKED EXAMPLE 2.17
The diagram shows the graph of y = x 2 and its image, y = g( x ), after a combination
of transformations.
y
y = x2
y = g(x)
4
3
62
SA
M
2
1
–2
–1
0
1
2
3
x
a Find two different ways of describing the combination of transformations.
b Write down the equation of the graph y = g( x ).
Answer
 4
1
a Translation of   followed by a horizontal stretch, stretch factor .
2
 0
y
y = x2
4
y = g(x)
3
2
1
–2
–1
6 x
0
1
2
3
4
5
6
x
Original material © Cambridge University Press 2017
Chapter 2: Functions
OR
Horizontal stretch, factor
 2
1
, followed by a translation of   .
2
 0
y
y = x2
y = g(x)
4
E
3
2
–3
PL
1
–2
0
–1
1
2
3
x
b Using the first combination of transformations:
TIP
 4
translation of   means ‘replace x by x − 4 ’
 0
y = x2
becomes
Horizontal stretch, factor
The same answer will
be obtained when
using the second
combination of
transformations. You
may wish to check this
yourself.
y = ( x − 4)2
1
means ‘replace x by 2x’
2
y = ( x − 4)2 becomes y = (2 x − 4)2
SA
M
Hence g( x ) = (2 x − 4)2.
EXERCISE 2H
1 The diagram shows the graph of y = g( x ) .
Sketch the graph of each of the following.
a
y = g( x + 2) + 3
b
y = 2g( x ) + 1
c
y = 2 − g( x )
d
y = 2g( − x ) + 1
e
y = −2g( x ) − 1
f
y = g(2 x ) + 3
g
y = g(2 x − 6)
h
y = g( − x + 1)
y
2
y = g(x)
1
–3 –2 –1 O
–1
1
2
3 x
–2
–3
Original material © Cambridge University Press 2017
63
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
y
2 The diagram shows the graph of y = f( x ).
2
Write down, in terms of f( x ), the equation of
the graph of each of the following diagrams.
y = f(x)
1
–4 –3 –2 –1 O
–1
1
2
3
x
–2
b
y
4
y
4
3
3
2
2
1
1
–4 –3 –2 –1 O
–1
1
2
3
4 x
–4 –3 –2 –1–1O
3
2
1
1
2
3
4 x
SA
M
–3
4 x
PL
y
4
–2
3
–4
–4
64
2
–3
–3
–4 –3 –2 –1–1O
1
–2
–2
c
E
a
–4
3 Given that y = x 2 find the image of the curve y = x 2 after each of the following
combinations of transformations:
a a stretch in the y-direction with factor 3 followed by a translation by the
 1
vector  
 0
 1
b a translation by the vector   followed by a stretch in the y-direction with
 0
factor 3.
4 Find the equation of the image of the curve y = x 2 after each of the following
combinations of transformations and in each case sketch the graph of the
resulting curve:
a a stretch in the x-direction with factor 2 followed by a translation by the
 5
vector  
 0
 5
b a translation by the vector   followed by a stretch in the x-direction with
 0
factor 2.
c
On a graph show the curve y = x 2 and each of your answers to parts a and b.
Original material © Cambridge University Press 2017
Chapter 2: Functions
5 Given that f( x ) = x 2 + 1 find the image of y = f( x ) after each of the following
combinations of transformations:
 0
a translation   , followed by a stretch parallel to the y-axis with stretch factor 2
 −5 
 2
b translation   , followed by a reflection in the x-axis.
 0
E
6 a The graph of y = g( x ) is reflected in the y-axis and then stretched with
stretch factor 2 parallel to the y-axis. Write down the equation of the
resulting graph.
 2
b The graph of y = f( x ) is translated by the vector   and then reflected in
 −3 
the x-axis. Write down the equation of the resulting graph.
PL
7 Determine the sequence of transformations that maps y = f( x ) to each of the
following functions.
1
a y = f ( x ) + 3 b y = −f( x ) + 2 c y = f(2 x − 6) d y = 2f( x ) − 8
2
8 Determine the sequence of transformations that maps:
1
a the curve y = x 3 onto the curve y = ( x + 5)3
2
1
b the curve y = x 3 onto the curve y = − ( x + 1)3 − 2
2
c the curve y = 3 x onto the curve y = −2 3 x − 3 + 4
9 Given that f( x ) =
x , write down the equation of the image of f( x ) after:
SA
M
 0
a reflection in the x-axis, followed by translation   , followed by translation
 3
 1
 0  , followed by a stretch parallel to the x-axis with stretch factor 2
 0
b translation   , followed by a stretch parallel to the x-axis with stretch
 3
 1
factor 2, followed by a reflection in the x-axis, followed by translation   .
 0
10 Given that g( x ) = x 2 , write down the equation of the image of g( x ) after:
 −4 
a translation   , followed by a reflection in the y-axis, followed by translation
 0
 0
 2  , followed by a stretch parallel to the y-axis with stretch factor 3
b a stretch parallel to the y-axis with stretch factor 3, followed by translation
 −4 
 0
 2  , followed by reflection in y-axis, followed by translation  0  .
PS
PS
11 Find two different ways of describing the combination of transformations that
maps the graph of f( x ) = x onto the graph g( x ) = − x − 2 and sketch the
graphs of y = f( x ) and y = g( x ).
12 Find two different ways of describing the sequence of transformations that maps
the graph of y = f( x ) onto the graph of y = f(2 x + 10) .
Original material © Cambridge University Press 2017
WEB LINK
Try the Transformers
resource on the
Underground
Mathematics website.
65
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
Checklist of learning and understanding
Functions
A function is a rule that maps each x value to just one y value for a defined set of input values.
●●
A function can be either one-one or many-one.
●●
The set of input values for a function is called the domain of the function.
●●
The set of output values for a function is called the range (or image set) of the function.
E
●●
Composite functions
fg( x ) means the function g acts on x first, then f acts on the result.
●●
fg only exists if the range of g is contained within the domain of f.
●●
In general, fg( x ) ≠ gf( x ).
Inverse functions
●●
PL
●●
The inverse of a function f( x ) is the function that undoes what f( x ) has done.
f f −1( x ) = f −1 f( x ) = x or: if y = f( x ) then x = f −1( y )
●●
The inverse of the function f( x ) is written as f −1 ( x ).
●●
The steps for finding the inverse function are:
Step 1: Write the function as y =
Step 2: Interchange the x and y variables.
Step 3: Rearrange to make y the subject.
●●
The domain of f −1 ( x ) is the range of f( x ).
●●
The range of f −1 ( x ) is the domain of f( x ).
●●
An inverse function f −1 ( x ) can exist if, and only if, the function f( x ) is one-one.
●●
The graphs of f and f −1 are reflections of each other in the line y = x.
●●
If f( x ) = f −1 ( x ), then the function f is called a self-inverse function.
●●
If f is self-inverse then ff( x ) = x.
●●
The graph of a self-inverse functions has y = x as a line of symmetry.
SA
M
66
Transformations of functions
●●
 0
The graph of y = f( x ) + a is a translation of y = f( x ) by the vector   .
 a
●●
The graph of y = f( x + a ) is a translation of y = f( x ) by the vector
●●
The graph of y = − f( x ) is a reflection of the graph y = f( x ) in the x-axis.
●●
The graph of y = f( − x ) is a reflection of the graph y = f( x ) in the y-axis.
 −a 
.
 0 


The graph of y = a f( x ) is a stretch of y = f( x ), stretch factor a, parallel to the y-axis.
1
●● The graph of y = f( ax ) is a stretch of y = f( x ), stretch factor , parallel to the x-axis.
a
●●
Combining transformations
●●
When two vertical transformations or two horizontal transformations are combined, the order
in which they are applied may affect the outcome.
●●
When one horizontal and one vertical transformation are combined, the order in which they are
applied does not affect the outcome.
●●
Vertical transformations follow the ‘normal’ order of operations as used in arithmetic
●●
Horizontal transformations follow the opposite order to the ‘normal’ order of operations as used
in arithmetic.
Original material © Cambridge University Press 2017
Chapter 2: Functions
END-OF-CHAPTER REVIEW EXERCISE 2
1
Functions f and g are defined for x ∈ by:
f : x 3x − 1
g : x 5x − x 2
Express gf( x ) in the form a − b( x − c )2, where a, b and c are constants.
y
E
2
2
3
x
PL
O
[5]
The diagram shows a sketch of the curve with equation y = f( x ).
1
a Sketch the graph of y = − f  x  .
2 
b Describe fully a sequence of two transformations that maps the graph of y = f( x ) onto the
graph of y = f(3 − x ) .
3
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 2
b The curve is translated by the vector   then stretched vertically with stretch factor 3.
 0
Find the equation of the resulting curve, giving your answer in the form y = ax 2 + bx .
[2]
67
[4]
The function f : x x 2 − 2 is defined for the domain x > 0 .
a Find f −1( x ) and state the domain of f −1.
5
[2]
A curve has equation y = x 2 + 6x + 8.
a Sketch the curve showing the coordinates of any axis crossing points.
4
[3]
[3]
−1
b On the same diagram, sketch the graphs of f and f .
[3]
i Express − x 2 + 6x − 5 in the form a ( x + b )2 + c, where a, b and c are constants.
[3]
The function f : x − x 2 + 6x − 5 is defined for x > m, where m is a constant.
ii State the smallest possible value of m for which f is one-one.
−1
[1]
−1
iii For the case where m = 5, find an expression for f ( x ) and state the domain of f .
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 November 2015
6
The function f : x x 2 − 4x + k is defined for the domain x > p , where k and p are constants.
i Express f( x ) in the form ( x + a )2 + b + k, where a and b are constants.
[2]
ii State the range of f in terms of k.
[1]
iii State the smallest value of p for which f is one-one.
[1]
iv F
or the value of p found in part iii, find an expression for f −1( x ) and state the domain of f −1,
giving your answer in terms of k.
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2012
Original material © Cambridge University Press 2017
Cambridge International AS & A Level Mathematics: Pure Mathematics 1
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x
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The diagram shows the function f defined for −1 < x < 4 , where
for − 1 < x < 1,
for 1 < x < 4.
PL
 3x − 2

f( x ) =  4
 5 − x
i State the range of f .
[1]
ii Copy the diagram and on your copy sketch the graph of y = f −1( x ) .
[2]
iii O
btain expressions to define the function f −1, giving also the set of values for which each
expression is valid.
[6]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 June 2014
8
The function f is defined by f( x ) = 4x 2 − 24x + 11 , for x ∈  .
i E
xpress f( x ) in the form a ( x − b )2 + c and hence state the coordinates of the vertex of the
graph of y = f( x ) .
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[4]
The function g is defined by g( x ) = 4x 2 − 24x + 11 , for x < 1.
ii State the range of g.
[2]
iii Find an expression for g −1( x ) and state the domain of g −1.
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2012
9
i Express 2 x 2 − 12 x + 13 in the form a ( x + b )2 + c , where a, b and c are constants.
[3]
2
ii T
he function f is defined by f( x ) = 2 x − 12 x + 13 , for x > k , where k is a constant. It is given
that f is a one-one function. State the smallest possible value of k.
[1]
The value of k is now given to be 7.
iii Find the range of f.
[1]
iv Find the expression for f −1( x ) and state the domain of f −1.
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q8 June 2013
10 i Express x 2 − 2 x − 15 in the form ( x + a )2 + b.
[2]
The function f is defined for p < x < q, where p and q are constants, by
f : x x 2 − 2 x − 15.
The range of f is given by c < f( x ) < d, where c and d are constants.
ii State the smallest possible value of c.
Original material © Cambridge University Press 2017
[1]
Chapter 2: Functions
For the case where c = 9 and d = 65,
iii find p and q,
[4]
iv find an expression for f −1( x ).
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q10 November 2014
i
Express f( x ) in the form a ( x − b )2 − c.
ii State the range of f .
E
11 The function f is defined by f : x 2 x 2 − 12 x + 7 for x ∈ .
[3]
[1]
iii Find the set of values of x for which f( x ) < 21.
[3]
PL
The function g is defined by g : x 2 x + k for x ∈ .
iv Find the value of the constant k for which the equation gf( x ) = 0 has two equal roots.
[4]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q9 June 2010
12 Functions f and g are defined for x ∈ by
f : x 2 x + 1,
g : x x 2 − 2.
i
Find and simplify expressions for fg( x ) and gf( x ).
[2]
[3]
iii Find the value of b ( b ≠ a ) for which g( b ) = b.
[2]
iv Find and simplify an expression for f −1g( x ) .
[2]
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ii Hence find the value of a for which fg( a ) = gf( a ).
The function h is defined by
h : x x 2 − 2, for x < 0 .
v Find an expression for h −1( x ) .
[2]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 June 2011
13 Functions f and g are defined for x ∈ by
f : x 2 x 2 − 8x + 10 for 0 < x < 2,
g:xx
i
for 0 < x < 10.
Express f( x ) in the form a ( x + b )2 + c, where a, b and c are constants.
[3]
ii State the range of f.
[1]
iii State the domain of f −1.
[1]
iv S
ketch on the same diagram the graphs of y = f( x ), y = g( x ) and y = f −1( x ), making clear
the relationships between the graphs.
[4]
v Find an expression for f −1( x ).
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q11 November 2011
Original material © Cambridge University Press 2017
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